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Chapter 13

Section and Mid-Point Formulae

Class - 10 RS Aggarwal Mathematics Solutions

Exercise 13

Question 1(i)

Find the co-ordinates of point P which divides the line segment joining A(-2, -7) and B(6, 1) in the ratio 5 : 3.

Answer

Let point P be (x, y).

Given,

m1 : m2 = 5 : 3

Find the co-ordinates of point P which divides the line segment joining A(-2, -7) and B(6, 1) in the ratio 5 : 3. Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒(x,y)=(5×6+3×−25+3,5×1+3×−75+3)⇒(x,y)=(30−68,5−218)⇒(x,y)=(248,−168)⇒(x,y)=(3,−2).\Rightarrow (x, y) = \Big(\dfrac{5 \times 6 + 3 \times -2}{5 + 3}, \dfrac{5 \times 1 + 3 \times -7}{5 + 3}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{30 - 6}{8}, \dfrac{5 - 21}{8}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{24}{8}, \dfrac{-16}{8}\Big) \\[1em] \Rightarrow (x, y) = (3, -2).

Hence, the coordinates of P are (3, -2).

Question 1(ii)

The line segment joining the points A(4, -3) and B(4, 2) is divided by the point P such that AP : AB = 2 : 5. Find the co-ordinates of P.

Answer

Let point P be (x, y).

The line segment joining the points A(4, -3) and B(4, 2) is divided by the point P such that AP : AB = 2 : 5. Find the co-ordinates of P. Reflection, RSA Mathematics Solutions ICSE Class 10.

⇒APAB=25⇒APAP+PB=25⇒5AP=2AP+2PB⇒3AP=2PB⇒APPB=23⇒AP:PB=2:3.\Rightarrow \dfrac{AP}{AB} = \dfrac{2}{5} \\[1em] \Rightarrow \dfrac{AP}{AP + PB} = \dfrac{2}{5} \\[1em] \Rightarrow 5AP = 2AP + 2PB \\[1em] \Rightarrow 3AP = 2PB \\[1em] \Rightarrow \dfrac{AP}{PB} = \dfrac{2}{3} \\[1em] \Rightarrow AP : PB = 2 : 3.

m1 : m2 = AP : PB = 2 : 3.

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒(x,y)=(2×4+3×42+3,2×2+3×−32+3)⇒(x,y)=(8+125,4−95)⇒(x,y)=(205,−55)⇒(x,y)=(4,−1).\Rightarrow (x, y) = \Big(\dfrac{2 \times 4+ 3 \times 4}{2 + 3}, \dfrac{2 \times 2 + 3 \times -3}{2 + 3}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{8 + 12}{5}, \dfrac{4 - 9}{5}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{20}{5}, \dfrac{-5}{5}\Big) \\[1em] \Rightarrow (x, y) = (4, -1).

Hence, the coordinates of P are (4, -1).

Question 2

P(1, -2) is a point on the line segment A(3, -6) and B(x, y) such that AP : PB is equal to 2 : 3. Find the co-ordinates of B.

Answer

Given,

m1 : m2 = 2 : 3

By section-formula,

x = (m1x2+m2x1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}\Big)

P(1, -2) is a point on the line segment A(3, -6) and B(x, y) such that AP : PB is equal to 2 : 3. Find the co-ordinates of B. Reflection, RSA Mathematics Solutions ICSE Class 10.

Substituting values we get :

⇒1=(2×x+3×32+3)⇒1=(2x+95)⇒5=2x+9⇒5−9=2x⇒−4=2x⇒x=−42⇒x=−2.\Rightarrow 1 = \Big(\dfrac{2 \times x + 3 \times 3}{2 + 3}\Big) \\[1em] \Rightarrow 1 = \Big(\dfrac{2x + 9}{5}\Big) \\[1em] \Rightarrow 5 = 2x + 9 \\[1em] \Rightarrow 5 - 9 = 2x \\[1em] \Rightarrow -4 = 2x \\[1em] \Rightarrow x = \dfrac{-4}{2} \\[1em] \Rightarrow x = -2.

By section-formula,

y = (m1y2+m2y1m1+m2)\Big(\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒−2=(2×y+3×−62+3)⇒−2=(2y−185)⇒−10=2y−18⇒−10+18=2y⇒8=2y⇒y=82⇒y=4.\Rightarrow -2 = \Big(\dfrac{2 \times y + 3 \times -6}{2 + 3}\Big) \\[1em] \Rightarrow -2 = \Big(\dfrac{2y - 18}{5}\Big) \\[1em] \Rightarrow -10 = 2y - 18 \\[1em] \Rightarrow -10 + 18 = 2y \\[1em] \Rightarrow 8 = 2y \\[1em] \Rightarrow y = \dfrac{8}{2} \\[1em] \Rightarrow y = 4.

Hence, the coordinates of B are (-2, 4).

Question 3

Find a point P on the line segment joining A(14,-5) and (-4,4), which is twice as far from A as from B

Answer

Let point P be (x, y).

P(1, -2) is a point on the line segment A(3, -6) and B(x, y) such that AP : PB is equal to 2 : 3. Find the co-ordinates of B. Reflection, RSA Mathematics Solutions ICSE Class 10.

Since, point P is twice as far from A as from B.

⇒ AP = 2BP

⇒ APBP=21\dfrac{AP}{BP} = \dfrac{2}{1}

⇒ AP : BP = 2 : 1.

⇒ m1 : m2 = AP : PB = 2 : 1

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒(x,y)=(2×−4+1×142+1,2×4+1×−52+1)⇒(x,y)=(−8+143,8−53)⇒(x,y)=(63,33)⇒(x,y)=(2,1).\Rightarrow (x, y) = \Big(\dfrac{2 \times -4+ 1 \times 14}{2 + 1}, \dfrac{2 \times 4 + 1 \times -5}{2 + 1}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{-8 + 14}{3}, \dfrac{8 - 5}{3}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{6}{3}, \dfrac{3}{3}\Big) \\[1em] \Rightarrow (x, y) = (2, 1).

Hence, the coordinates of P are (2, 1).

Question 4

Find the co-ordinates of the points of trisection of the line segment joining the points A(5, -3) and B(2, -9).

Answer

Let point P is the first point of trisection, meaning it divides the segment AB internally in the ratio m1 : m2 = 1 : 2

Find the co-ordinates of the points of trisection of the line segment joining the points A(5, -3) and B(2, -9). Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒P(x,y)=(1×2+2×51+2,1×−9+2×−31+2)⇒P(x,y)=(2+103,−9−63)⇒P(x,y)=(123,−153)⇒P(x,y)=(4,−5).\Rightarrow P(x, y) = \Big(\dfrac{1 \times 2 + 2 \times 5}{1 + 2}, \dfrac{1 \times -9 + 2 \times -3}{1 + 2}\Big) \\[1em] \Rightarrow P(x, y) = \Big(\dfrac{2 + 10}{3}, \dfrac{-9 - 6}{3}\Big) \\[1em] \Rightarrow P(x, y) = \Big(\dfrac{12}{3}, \dfrac{-15}{3}\Big) \\[1em] \Rightarrow P(x, y) = (4, -5).

Let point Q is the second point of trisection, meaning it divides the segment AB internally in the ratio m1 : m2 = 2 : 1

Substituting values we get :

⇒Q(a,b)=(2×2+1×52+1,2×−9+1×−32+1)⇒Q(a,b)=(4+53,−18−33)⇒Q(a,b)=(93,−213)⇒Q(a,b)=(3,−7).\Rightarrow Q(a, b) = \Big(\dfrac{2 \times 2 + 1 \times 5}{2 + 1}, \dfrac{2 \times -9 + 1 \times -3}{2 + 1}\Big) \\[1em] \Rightarrow Q(a, b) = \Big(\dfrac{4 + 5}{3}, \dfrac{-18 - 3}{3}\Big) \\[1em] \Rightarrow Q(a, b) = \Big(\dfrac{9}{3}, \dfrac{-21}{3}\Big) \\[1em] \Rightarrow Q(a, b) = (3, -7).

Hence, the coordinates of trisection are P(4, -5) and Q(3, -7).

Question 5

Find the co-ordinates of the mid-point of the line segment joining :

(i) A(5, 7) and B(-3, -1)

(ii) P(-5, -8) and Q(3, 4)

Answer

(i) Let mid-point of the line segment joining AB be P(x,y).

By using mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Substituting values we get :

⇒(x,y)=(5+(−3)2,7+(−1)2)⇒(x,y)=(22,62)⇒(x,y)=(1,3).\Rightarrow (x, y) = \Big(\dfrac{5 + (-3)}{2}, \dfrac{7 + (-1)}{2}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{2}{2}, \dfrac{6}{2}\Big) \\[1em] \Rightarrow (x, y) = (1, 3).

Hence, the coordinates of mid-point are (1, 3).

(ii) Let mid-point of the line segment joining PQ be B(x,y).

By using mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Substituting values we get :

⇒(x,y)=(−5+32,−8+42)⇒(x,y)=(−22,−42)⇒(x,y)=(−1,−2).\Rightarrow (x, y) = \Big(\dfrac{-5 + 3}{2}, \dfrac{-8 + 4}{2}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{-2}{2}, \dfrac{-4}{2}\Big) \\[1em] \Rightarrow (x, y) = (-1, -2).

Hence, the coordinates of mid-point are (-1, -2).

Question 6

The line segment joining A(-3, 1) and B(7, -5) is a diameter of a circle whose centre is C. Find the co-ordinates of the centre C.

Answer

Since the line segment joining points A and B is the diameter of the circle, the centre of the circle must be the mid-point of the diameter AB.

The line segment joining A(-3, 1) and B(7, -5) is a diameter of a circle whose centre is C. Find the co-ordinates of the centre C. Reflection, RSA Mathematics Solutions ICSE Class 10.

By using mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Substituting values we get :

⇒C(x,y)=(−3+72,1+(−5)2)⇒C(x,y)=(42,−42)⇒C(x,y)=(2,−2).\Rightarrow C(x, y) = \Big(\dfrac{-3 + 7}{2}, \dfrac{1 + (-5)}{2}\Big) \\[1em] \Rightarrow C(x, y) = \Big(\dfrac{4}{2}, \dfrac{-4}{2}\Big) \\[1em] \Rightarrow C(x, y) = (2, -2).

Hence, the coordinates of centre (C) are (2, -2).

Question 7

A(10, 5), B(6, -3) and C(2, 1) are the vertices of a ΔABC. L is the mid-point of AB and M is the mid-point of AC. Write down the co-ordinates of L and M. Show that LM=12BC.LM = \dfrac{1}{2} BC.

Answer

By using mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

L is the mid-point of AB.

A(10, 5), B(6, -3) and C(2, 1) are the vertices of a ΔABC. L is the mid-point of AB and M is the mid-point of AC. Write down the co-ordinates of L and M. Show that. Reflection, RSA Mathematics Solutions ICSE Class 10.

Substituting values we get :

⇒L=(10+62,5+(−3)2)⇒L=(162,22)⇒L=(8,1).\Rightarrow L = \Big(\dfrac{10 + 6}{2}, \dfrac{5 + (-3)}{2}\Big) \\[1em] \Rightarrow L = \Big(\dfrac{16}{2}, \dfrac{2}{2}\Big) \\[1em] \Rightarrow L = (8, 1).

Given,

M is the mid-point of AC.

Substituting values we get :

⇒M=(10+22,5+12)⇒M=(122,62)⇒M=(6,3).\Rightarrow M = \Big(\dfrac{10 + 2}{2}, \dfrac{5 + 1}{2}\Big) \\[1em] \Rightarrow M = \Big(\dfrac{12}{2}, \dfrac{6}{2}\Big) \\[1em] \Rightarrow M = (6, 3).

By distance formula,

d = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Substituting values we get :

LM=(6−8)2+(3−1)2=(−2)2+(2)2=4+4=8=22.BC=(2−6)2+(1−(−3))2=(−4)2+(4)2=16+16=32=16×2=42=2×22=2×LM.LM = \sqrt{(6 - 8)^2 + (3 - 1)^2} \\[1em] = \sqrt{(-2)^2 + (2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} = 2\sqrt{2}. \\[1em] BC = \sqrt{(2 - 6)^2 + (1 - (-3))^2} \\[1em] = \sqrt{(-4)^2 + (4)^2} \\[1em] = \sqrt{16 + 16} \\[1em] = \sqrt{32} \\[1em] = \sqrt{16 \times 2} \\[1em] = 4\sqrt{2} \\[1em] = 2 \times 2\sqrt{2} \\[1em] = 2 \times LM.

Thus, BC = 2LM or LM = 12BC.\dfrac{1}{2}BC.

Hence, proved that LM=12BCLM = \dfrac{1}{2}BC.

Question 8

The mid-point of the line segment joining A(p, 5) and B(3, q) is M(-1, 4). Find the values of p and q.

Answer

Given,

Mid-point of the line segment joining A(p, 5) and B(3, q) is M(-1, 4).

The mid-point of the line segment joining A(p, 5) and B(3, q) is M(-1, 4). Find the values of p and q. Reflection, RSA Mathematics Solutions ICSE Class 10.

By using mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Substituting values we get :

⇒(−1,4)=(p+32,5+q2)⇒(−1,4)=(p+32,5+q2)⇒−1=p+32 and 4=5+q2⇒p+3=−2 and 5+q=8⇒p=−2−3 and q=8−5⇒p=−5 and q=3.\Rightarrow (-1, 4) = \Big(\dfrac{p + 3}{2}, \dfrac{5 + q}{2} \Big) \\[1em] \Rightarrow (-1, 4) = \Big(\dfrac{p + 3}{2}, \dfrac{5 + q}{2}\Big) \\[1em] \Rightarrow -1 = \dfrac{p + 3}{2} \text{ and } 4 = \dfrac{5 + q}{2} \\[1em] \Rightarrow p + 3 = -2 \text{ and } 5 + q = 8 \\[1em] \Rightarrow p = -2 - 3 \text{ and } q = 8 - 5 \\[1em] \Rightarrow p = -5 \text{ and } q = 3.

Hence, p = -5 and q = 3.

Question 9

The centre of a circle is C(-2, 3) and one end of a diameter PQ is P(2, -4). Find the co-ordinates of Q.

Answer

Since PQ is the diameter of the circle and C is the center, C must be the mid-point of the segment PQ.

Let coordinates of Q be (x1, y1).

The centre of a circle is C(-2, 3) and one end of a diameter PQ is P(2, -4). Find the co-ordinates of Q. Reflection, RSA Mathematics Solutions ICSE Class 10.

By using mid-point formula,

x-coordinate = (x1+x22)\Big(\dfrac{x_1 + x_2}{2}\Big)

Substituting values we get :

⇒−2=(2+x12)⇒−4=2+x1⇒−4−2=x1⇒x1=−6.\Rightarrow -2 = \Big(\dfrac{2 + x_1}{2}\Big) \\[1em] \Rightarrow -4 = 2 + x_1 \\[1em] \Rightarrow -4 - 2 = x_1 \\[1em] \Rightarrow x_1 = -6.

By using mid-point formula,

y-coordinate = (y1+y22)\Big(\dfrac{y_1 + y_2}{2}\Big)

Substituting values we get :

⇒3=(−4+y12)⇒6=−4+y1⇒6+4=y1⇒y1=10.\Rightarrow 3 = \Big(\dfrac{-4 + y_1}{2}\Big) \\[1em] \Rightarrow 6 = -4 + y_1 \\[1em] \Rightarrow 6 + 4 = y_1 \\[1em] \Rightarrow y_1 = 10.

Q = (x1, y1) = (-6, 10).

Hence, coordinates of Q are (-6, 10).

Question 10

The point P(-4, 1) divides the line segment joining the points A(2, -2) and B in the ratio 3 : 5. Find the co-ordinates of point B.

Answer

Let coordinates of B be (a, b).

Point P(-4, 1) divides the line segment joining A(2, -2) and B(a, b) in the ratio 3 : 5.

The point P(-4, 1) divides the line segment joining the points A(2, -2) and B in the ratio 3 : 5. Find the co-ordinates of point B. Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒(−4,1)=(3a+5(2)3+5,3b+5(−2)3+5)⇒(−4,1)=(3a+108,3b−108)⇒−4=(3a+108) and 1=(3b−108)⇒−32=3a+10 and 8=3b−10⇒−32−10=3a and 8+10=3b⇒−42=3a and 18=3b⇒a=−423 and b=183⇒a=−14 and b=6.\Rightarrow (-4, 1) = \Big(\dfrac{3a + 5(2)}{3 + 5}, \dfrac{3b + 5(-2)}{3 + 5}\Big) \\[1em] \Rightarrow (-4, 1) = \Big(\dfrac{3a + 10}{8}, \dfrac{3b - 10}{8}\Big) \\[1em] \Rightarrow -4 = \Big(\dfrac{3a + 10}{8}\Big) \text{ and } 1 = \Big(\dfrac{3b - 10}{8}\Big) \\[1em] \Rightarrow -32 = 3a + 10 \text{ and } 8 = 3b - 10 \\[1em] \Rightarrow -32 - 10 = 3a \text{ and } 8 + 10 = 3b \\[1em] \Rightarrow -42 = 3a \text{ and } 18 = 3b \\[1em] \Rightarrow a = \dfrac{-42}{3} \text{ and } b = \dfrac{18}{3} \\[1em] \Rightarrow a = -14 \text{ and }b = 6 .

B = (a, b) = (-14, 6).

Hence, coordinates of B(-14, 6).

Question 11

In what ratio does the point P(2, -5) divide the join of A(-3, 5) and B(4, -9)?

Answer

Let the point P(2, -5) divide the line segment AB in the ratio k : 1.

In what ratio does the point P(2, -5) divide the join of A(-3, 5) and B(4, -9)? Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula,

x-coordinate = (m1x2+m2x1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒2=(k(4)+1(−3)k+1)⇒2(k+1)=4k−3⇒2k+2=4k−3⇒2+3=4k−2k⇒5=2k⇒k=52⇒k:1=52:1=5:2.\Rightarrow 2 = \Big(\dfrac{k(4) + 1(-3)}{k + 1}\Big) \\[1em] \Rightarrow 2(k + 1) = 4k - 3 \\[1em] \Rightarrow 2k + 2 = 4k - 3 \\[1em] \Rightarrow 2 + 3 = 4k - 2k \\[1em] \Rightarrow 5 = 2k \\[1em] \Rightarrow k = \dfrac{5}{2} \\[1em] \Rightarrow k : 1 = \dfrac{5}{2} : 1 = 5 : 2.

Hence, point P divide AB in the ratio 5 : 2.

Question 12

In what ratio does the point P(a, -1) divide the join of A(1, -3) and B(6, 2)? Hence, find the value of a.

Answer

Let the point P(2, -5) divide the segment AB in the ratio k : 1.

In what ratio does the point P(a, -1) divide the join of A(1, -3) and B(6, 2)? Hence, find the value of a. Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula,

y = (m1y2+m2y1m1+m2)\Big(\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒−1=(k(2)+1(−3)k+1)⇒−1(k+1)=2k−3⇒−k−1=2k−3⇒−1+3=2k+k⇒3k=2⇒k=23⇒k:1=23:1=2:3.\Rightarrow -1 = \Big(\dfrac{k(2) + 1(-3)}{k + 1}\Big) \\[1em] \Rightarrow -1(k + 1) = 2k - 3 \\[1em] \Rightarrow -k - 1 = 2k - 3 \\[1em] \Rightarrow -1 + 3 = 2k + k \\[1em] \Rightarrow 3k = 2 \\[1em] \Rightarrow k = \dfrac{2}{3} \\[1em] \Rightarrow k : 1 = \dfrac{2}{3} : 1 = 2 : 3.

By section-formula,

x-coordinate = (m1x2+m2x1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}\Big)

Substitute values we get:

⇒a=(2×6+3×12+3)=(12+35)=155=3.\Rightarrow a = \Big(\dfrac{2 \times 6 + 3 \times 1}{2 + 3}\Big) \\[1em] = \Big(\dfrac{12 + 3}{5}\Big) \\[1em] = \dfrac{15}{5} \\[1em] = 3.

Hence, point P divide AB in the ratio 2 : 3 and value of a = 3.

Question 13

The line segment joining A(2, 3) and B(6, -5) is intercepted by the x-axis at the point k. Find the ratio in which k divides AB. Also, write the co-ordinates of the point k.

Answer

Since the point k lies on the x-axis, its y-coordinate must be 0. Let the coordinates of k be (x, 0).

Let k divide the line segment joining A(2, 3) and B(6, -5) in the ratio m1:m2

The line segment joining A(2, 3) and B(6, -5) is intercepted by the x-axis at the point k. Find the ratio in which k divides AB. Also, write the co-ordinates of the point k. Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula,

y = (m1y2+m2y1m1+m2)\Big(\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒0=(m1(−5)+m2(3)m1+m2)⇒−5m1+3m2=0⇒5m1=3m2⇒m1m2=35⇒m1:m2=3:5.\Rightarrow 0 = \Big(\dfrac{m_1(-5) + m_2(3)}{m_1 + m_2}\Big) \\[1em] \Rightarrow -5m_1 + 3m_2 = 0 \\[1em] \Rightarrow 5m_1 = 3m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{3}{5} \\[1em] \Rightarrow m_1 : m_2 = 3 : 5.

By section-formula,

x = (m1x2+m2x1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒x=(3(6)+5(2)3+5)⇒x=(18+108)⇒x=(288)⇒x=72\Rightarrow x = \Big(\dfrac{3(6) + 5(2)}{3 + 5}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{18 + 10}{8}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{28}{8}\Big) \\[1em] \Rightarrow x = \dfrac{7}{2}

Hence, ratio in which k divides AB = 3 : 5 and coordinates of the point k are (72,0)\Big(\dfrac{7}{2}, 0\Big) .

Question 14

In what ratio is the segment joining the points A(6, 5) and B(-3, 2) divided by the y-axis? Find the point at which the y-axis cuts AB.

Answer

When a point lies on the y-axis, its x-coordinate is always 0. Let the point where the y-axis cuts AB be P(0, y).

Let ratio in which P divides AB be m1 : m2.

In what ratio is the segment joining the points A(6, 5) and B(-3, 2) divided by the y-axis? Find the point at which the y-axis cuts AB. Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula,

x = (m1x2+m2x1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒0=(m1(−3)+m2(6)m1+m2)⇒0=−3m1+6m2⇒3m1=6m2⇒m1m2=63=2.\Rightarrow 0 = \Big(\dfrac{m_1(-3) + m_2(6)}{m_1 + m_2}\Big) \\[1em] \Rightarrow 0 = -3m_1 + 6m_2 \\[1em] \Rightarrow 3m_1 = 6m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{6}{3} = 2.

Thus, m1 : m2 = 2 : 1.

By section-formula,

y = (m1y2+m2y1m1+m2)\Big(\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substitute values we get:

⇒y=(2(2)+1(5)2+1)=(4+53)=(93)=3.\Rightarrow y = \Big(\dfrac{2(2) + 1(5)}{2 + 1}\Big) \\[1em] = \Big(\dfrac{4 + 5}{3}\Big) \\[1em] = \Big(\dfrac{9}{3}\Big) \\[1em] = 3.

P = (0, y) = (0, 3).

Hence, AB is divided in ratio 2 : 1 and point at which the y-axis cuts AB is (0, 3).

Question 15

(i) Write down the co-ordinates of the point P that divides the line segment joining A(-4, 1) and B(17, 10) in the ratio 1 : 2.

(ii) Calculate the distance OP, where O is the origin.

(iii) In what ratio does the y-axis divide the line AB?

Answer

(i) The point P divides the line segment joining A(-4, 1) and B(17, 10) in the ratio m1 : m2 = 1 : 2.

Let coordinates of P be (x, y).

Draw co-ordinate axes and represent the following points : Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒(x,y)=(1(17)+2(−4)1+2,1(10)+2(1)1+2)⇒(x,y)=(17−83,10+23)⇒(x,y)=(93,123)⇒(x,y)=(3,4).\Rightarrow (x, y) = \Big(\dfrac{1(17) + 2(-4)}{1 + 2}, \dfrac{1(10) + 2(1)}{1 + 2}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{17 - 8}{3}, \dfrac{10 + 2}{3}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{9}{3}, \dfrac{12}{3}\Big) \\[1em] \Rightarrow (x, y) = (3, 4).

P = (x, y) = (3, 4)

Hence, coordinates of P = (3, 4).

(ii) Using distance formula,

d = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Substitute values we get,

OP=(3−0)2+(4−0)2=(3)2+(4)2=9+16=25=5 units.OP = \sqrt{(3 - 0)^2 + (4 - 0)^2} \\[1em] = \sqrt{(3)^2 + (4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = \text{5 units}.

Hence, distance of OP is 5 units.

(iii) A point on the y-axis has an x-coordinate of 0. Let the y-axis cut AB at K(0, y). let the ratio be k : 1.

By section-formula,

x = (m1x2+m2x1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}\Big)

Substitute values we get,

⇒0=(k(17)+1(−4)k+1)⇒17k−4=0⇒17k=4⇒k=417⇒k:1=417:1=4:17.\Rightarrow 0 = \Big(\dfrac{k(17) + 1(-4)}{k + 1}\Big) \\[1em] \Rightarrow 17k - 4 = 0 \\[1em] \Rightarrow 17k = 4 \\[1em] \Rightarrow k = \dfrac{4}{17} \\[1em] \Rightarrow k : 1 = \dfrac{4}{17} : 1 = 4 : 17.

Hence, y-axis cut line AB in ratio 4 : 17.

Question 16

The line segment joining P(-4, 5) and Q(3, 2) intersects the y-axis at R. PM and QN are perpendiculars from P and Q on the x-axis. Find :

The line segment joining P(-4, 5) and Q(3, 2) intersects the y-axis at R. PM and QN are perpendiculars from P and Q on the x-axis. Find : Reflection, RSA Mathematics Solutions ICSE Class 10.

(i) the ratio PR : RQ

(ii) the coordinates of R.

(iii) the area of quadrilateral PMNQ.

Answer

(i) The point R lies on the y-axis, so its x-coordinate is 0.

The line segment joining P(-4, 5) and Q(3, 2) intersects the y-axis at R. PM and QN are perpendiculars from P and Q on the x-axis. Find : Reflection, RSA Mathematics Solutions ICSE Class 10.

Let R(0, y) divide the line segment PQ in the ratio m1 : m2

By section-formula,

x = (m1x2+m2x1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}\Big)

Substitute values we get,

⇒0=(m1(3)+m2(−4)m1+m2)⇒0=3m1−4m2⇒3m1=4m2⇒m1m2=43.\Rightarrow 0 = \Big(\dfrac{m_1(3) + m_2(-4)}{m_1 + m_2}\Big) \\[1em] \Rightarrow 0 = 3m_1 - 4m_2 \\[1em] \Rightarrow 3m_1 = 4m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{4}{3}.

Hence, the ratio PR : RQ = 4 : 3.

(ii) Given,

P(-4, 5) and Q(3, 2)

m1 : m2 = 4 : 3.

By section-formula,

y = (m1y2+m2y1m1+m2)\Big(\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substitute values we get,

⇒y=(4(2)+3(5)4+3)⇒y=(8+157)⇒y=(237).\Rightarrow y = \Big(\dfrac{4(2) + 3(5)}{4 + 3}\Big) \\[1em] \Rightarrow y = \Big(\dfrac{8 + 15}{7}\Big) \\[1em] \Rightarrow y = \Big(\dfrac{23}{7}\Big).

Hence, the coordinates of R = (0,237)\Big(0, \dfrac{23}{7}\Big).

(iii) From figure,

PQNM is a trapezium, with PM and QN being the parallel sides, as both are perpendicular to x-axis.

From figure,

PM = 5 units, MN = 7 units and QN = 2 units.

The area of a trapezoid is given by :

Area = 12\dfrac{1}{2} × Sum of parallel sides × Distance between them

=12×(PM+QN)×MN=12×(5+2)×7=492=24.5 sq. units= \dfrac{1}{2} \times (PM + QN) \times MN \\[1em] = \dfrac{1}{2} \times (5 + 2) \times 7 \\[1em] = \dfrac{49}{2} \\[1em] = 24.5 \text{ sq. units}

Hence, area of PQNM = 24.5 sq.units.

Question 17

In the given figure, the line segment AB meets x-axis at A and y-axis at B. The point P(-3, 1) on AB divides it in ratio 2 : 3. Find the coordinates of A and B.

In the given figure, the line segment AB meets x-axis at A and y-axis at B. The point P(-3, 1) on AB divides it in ratio 2 : 3. Find the coordinates of A and B. Reflection, RSA Mathematics Solutions ICSE Class 10.

Answer

Since, point A and B lies on x-axis and y-axis respectively. Let their coordinates be A(a, 0) and B(0, b).

Given,

The line segment AB be divided by point P(-3, 1) in the ratio AP : PB = 2 : 3.

By section formula,

⇒(x,y)=(m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)⇒(−3,1)=(2(0)+3(a)2+3,2(b)+3(0)2+3)⇒(−3,1)=(3a5,2b5)⇒−3=(3a5),1=(2b5)⇒−15=3a,5=2b⇒a=−153,b=52⇒a=−5,b=52⇒A=(a,0)=(−5,0)⇒B=(0,b)=(0,52).\Rightarrow (x, y) = \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] \Rightarrow (-3, 1) = \Big(\dfrac{2(0) + 3(a)}{2 + 3}, \dfrac{2(b) + 3(0)}{2 + 3}\Big) \\[1em] \Rightarrow (-3, 1) = \Big(\dfrac{3a}{5}, \dfrac{2b}{5}\Big) \\[1em] \Rightarrow -3 = \Big(\dfrac{3a}{5}\Big), 1 = \Big(\dfrac{2b}{5}\Big) \\[1em] \Rightarrow -15 = 3a, 5 = 2b \\[1em] \Rightarrow a = \dfrac{-15}{3}, b = \dfrac{5}{2} \\[1em] \Rightarrow a = -5, b = \dfrac{5}{2} \\[1em] \Rightarrow A = (a, 0) = (-5, 0) \\[1em] \Rightarrow B = (0, b) = \Big(0, \dfrac{5}{2}\Big).

Hence, A(-5, 0) and B(0,52)B\Big(0, \dfrac{5}{2}\Big).

Question 18

Show that the line segment joining the points A(-5, 8) and B(10, -4) is trisected by the coordinate axes. Also, find the points of trisection of AB.

Answer

Let x-axis divide AB in the ratio k : 1 at the point P(x, 0).

Show that the line segment joining the points A(-5, 8) and B(10, -4) is trisected by the coordinate axes. Also, find the points of trisection of AB. Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula,

y = (m1y2+m2y1m1+m2)\Big(\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get,

⇒0=(k(−4)+1(8)k+1)⇒0=(−4k+8k+1)⇒0=−4k+8⇒4k=8⇒k=84=2.\Rightarrow 0 = \Big(\dfrac{k(-4) + 1(8)}{k + 1}\Big) \\[1em] \Rightarrow 0 = \Big(\dfrac{-4k + 8}{k + 1}\Big) \\[1em] \Rightarrow 0 = -4k + 8 \\[1em] \Rightarrow 4k = 8 \\[1em] \Rightarrow k = \dfrac{8}{4} = 2.

The x-axis divides AB in the ratio k : 1 = 2 : 1.

Thus, m1 : m2 = 2 : 1.

By section-formula,

x = (m1x2+m2x1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}\Big)

Substituting values we get,

⇒x=(2(10)+1(−5)2+1)⇒x=(20−53)⇒x=(153)⇒x=5.\Rightarrow x = \Big(\dfrac{2(10) + 1(-5)}{2 + 1}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{20 - 5}{3}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{15}{3}\Big) \\[1em] \Rightarrow x = 5.

The coordinates of P = (x, 0) = (5, 0).

Let the y-axis divide AB in the ratio p : 1 at the point Q(0, y).

By section-formula,

x = (m1x2+m2x1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}\Big)

Substituting values we get,

⇒0=(p(10)+1(−5)p+1)⇒0=(10p−5p+1)⇒10p−5=0⇒10p=5⇒p=510⇒p=12⇒p:1=12:1=1:2.\Rightarrow 0 = \Big(\dfrac{p(10) + 1(-5)}{p + 1}\Big) \\[1em] \Rightarrow 0 = \Big(\dfrac{10p - 5}{p + 1}\Big) \\[1em] \Rightarrow 10p - 5 = 0 \\[1em] \Rightarrow 10p = 5 \\[1em] \Rightarrow p = \dfrac{5}{10} \\[1em] \Rightarrow p = \dfrac{1}{2} \\[1em] \Rightarrow p : 1 = \dfrac{1}{2} : 1 = 1 : 2.

By section-formula,

y = (m1y2+m2y1m1+m2)\Big(\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substitute values we get,

⇒y=(1×−4+2×81+2)=(−4+163)=123=4.\Rightarrow y = \Big(\dfrac{1 \times -4 + 2 \times 8}{1 + 2}\Big) \\[1em] = \Big(\dfrac{-4 + 16}{3}\Big) \\[1em] = \dfrac{12}{3} \\[1em] = 4.

The coordinates of Q = (0, y) = (0, 4).

Hence, points of trisection of AB are Q(5, 0) and P(0, 4).

Question 19

The mid-points of the sides BC, CA and AB of ΔABC are D(2, 1), E(-1, -3) and F(4, 5) respectively. Find the co-ordinates of A, B and C.

Answer

Let the vertices of ΔABC be A(x1, y1), B(x2, y2), and C(x3, y3).

The mid-points of the sides BC, CA and AB of ΔABC are D(2, 1), E(-1, -3) and F(4, 5) respectively. Find the co-ordinates of A, B and C. Reflection, RSA Mathematics Solutions ICSE Class 10.

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Given,

D(2, 1) is the midpoint of BC.

Substitute values we get,

⇒x2+x32=2⇒x2+x3=4 ....(1)⇒y2+y32=1⇒y2+y3=2 ....(2)\Rightarrow \dfrac{x_2 + x_3}{2} = 2 \\[1em] \Rightarrow x_2 + x_3 = 4 \text{ ....(1)} \\[1em] \Rightarrow \dfrac{y_2 + y_3}{2} = 1 \\[1em] \Rightarrow y_2 + y_3 = 2 \text{ ....(2)}

Given,

E(-1, -3) is the midpoint of CA.

⇒x3+x12=−1⇒x3+x1=−2 .....(3)⇒y3+y12=−3⇒y3+y1=−6 ....(4)\Rightarrow \dfrac{x_3 + x_1}{2} = -1 \\[1em] \Rightarrow x_3 + x_1 = -2 \text{ .....(3)} \\[1em] \Rightarrow \dfrac{y_3 + y_1}{2} = -3 \\[1em] \Rightarrow y_3 + y_1 = -6 \text{ ....(4)}

Given,

F(4, 5) is the midpoint of CA.

⇒x1+x22=4⇒x1+x2=8 ....(5)⇒y1+y22=5⇒y1+y2=10 ....(6)\Rightarrow \dfrac{x_1 + x_2}{2} = 4 \\[1em] \Rightarrow x_1 + x_2 = 8 \text{ ....(5)} \\[1em] \Rightarrow \dfrac{y_1 + y_2}{2} = 5 \\[1em] \Rightarrow y_1 + y_2 = 10 \text{ ....(6)}

Adding the three equations (1), (3) and (5), we get :

⇒ (x2 + x3) + (x3 + x1) + (x1 + x2) = 4 + (-2) + 8

⇒ 2x1 +2x2 + 2x3 = 10

⇒ 2(x1 +x2 + x3) = 10

⇒ (x1 +x2 + x3) = 102\dfrac{10}{2}

⇒ (x1 +x2 + x3) = 5 .....(7)

Subtract (Eq. 3) from (Eq. 7) :

⇒ (x1 +x2 + x3) - ( x2 + x3) = 5 - 4

⇒ (x1 +x2 + x3 -x2 - x3) = 5 - 4

⇒ x1 = 1.

Subtract (Eq. 2) from (Eq. 7) :

⇒ (x1 +x2 + x3) - ( x3 + x1) = 5 - (-2)

⇒ (x1 +x2 + x3 -x3 - x1) = 5 + 2

⇒ x2 = 7.

Subtract (Eq. 5) from (Eq. 7):

⇒ (x1 +x2 + x3) - ( x1 + x2) = 5 - 8

⇒ (x1 +x2 + x3 -x1 - x2) = -3

⇒ x3 = -3.

Adding equations (2), (4) and (5), we get :

⇒ (y2 + y3) + (y3 + y1) + (y1 + y2) = 2 + (-6) + 10

⇒ 2y1 +2y2 + 2y3 = 6

⇒ 2(y1 +y2 + y3) = 6

⇒ (y1 +y2 + y3) = 62\dfrac{6}{2}

⇒ (y1 +y2 + y3) = 3 .....(8)

Subtract (Eq. 2) from (Eq. 8):

⇒ (y1 +y2 + y3) - ( y2 + y3) = 3 - 2

⇒ (y1 +y2 + y3 -y2 - y3) = 1

⇒ y1 = 1.

Subtract (Eq. 4) from (Eq. 8):

⇒ (y1 + y2 + y3) - ( y3 + y1) = 3-(-6)

⇒ (y1 +y2 + y3 -y3 - y1) = 3 + 6

⇒ y2 = 9.

Subtract (Eq. 6) from (Eq. 8):

⇒ (y1 + y2 + y3) - ( y1 + y2) = 3 - 10

⇒ (y1 +y2 + y3 -y1 - y2) = -7

⇒ y3 = -7.

⇒ A = (x1, y1) = (1, 1), B = (x2, y2) = (7, 9), C = (x3, y3) = (-3, -7).

Hence, A = (1, 1), B = (7, 9) and C = (-3, -7).

Question 20

Prove that the points A(-2, -1), B(1, 0), C(4, 3) and D(1, 2) are the vertices of a parallelogram ABCD.

Answer

Given,

A(-2, -1), B(1, 0), C(4, 3) and D(1, 2).

By using mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

The Midpoint of Diagonal AC :

⇒M(AC)=(−2+42,−1+32)⇒M(AC)=(22,22)⇒M(AC)=(1,1).\Rightarrow M_{(AC)} = \Big(\dfrac{-2 + 4}{2}, \dfrac{-1 + 3}{2}\Big) \\[1em] \Rightarrow M_{(AC)} = \Big(\dfrac{2}{2}, \dfrac{2}{2}\Big) \\[1em] \Rightarrow M_{(AC)} = (1, 1).

The Midpoint of Diagonal BD :

⇒M(BD)=(1+12,0+22)⇒M(BD)=(22,22)⇒M(BD)=(1,1)\Rightarrow M_{(BD)} = \Big(\dfrac{1 + 1}{2}, \dfrac{0 + 2}{2}\Big) \\[1em] \Rightarrow M_{(BD)} = \Big(\dfrac{2}{2}, \dfrac{2}{2}\Big) \\[1em] \Rightarrow M_{(BD)} = (1, 1)

Since the midpoint of diagonal AC, MAC(1, 1), is the same as the midpoint of diagonal BD, MBD (1, 1), the diagonals AC and BD bisect each other.

We know that,

A quadrilateral is a parallelogram if its diagonals bisect each other.

Hence, proved that ABCD is a parallelogram.

Question 21

If the points A(-2, -1), B(1, 0), C(a, 3) and D(1, b) form a parallelogram, find the values of a and b.

Answer

We know that,

Diagonals of a parallelogram bisect each other.

Since, ABCD is a // gm.

Thus, Mid-point of AC = Mid-point of BD.

Given,

A(-2, -1), B(1, 0), C(a, 3) and D(1, b)

If the points A(-2, -1), B(1, 0), C(a, 3) and D(1, b) form a parallelogram, find the values of a and b. Reflection, RSA Mathematics Solutions ICSE Class 10.

By using mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Mid-point of AC :

⇒MAC=(−2+a2,−1+32)⇒MAC=(−2+a2,22)⇒MAC=(−2+a2,1).\Rightarrow M_{AC} = \Big(\dfrac{-2 + a}{2}, \dfrac{-1 + 3}{2}\Big) \\[1em] \Rightarrow M_{AC} = \Big(\dfrac{-2 + a}{2}, \dfrac{2}{2}\Big) \\[1em] \Rightarrow M_{AC} = \Big(\dfrac{-2 + a}{2}, 1\Big).

Midpoint of Diagonal BD :

⇒MBD=(1+12,0+b2)⇒MBD=(22,b2)⇒MBD=(1,b2).\Rightarrow M_{BD} = \Big(\dfrac{1 + 1}{2}, \dfrac{0 + b}{2}\Big) \\[1em] \Rightarrow M_{BD} = \Big(\dfrac{2}{2}, \dfrac{b}{2}\Big) \\[1em] \Rightarrow M_{BD} = \Big(1, \dfrac{b}{2}\Big) .

Equating the x-coordinates of mid-points of AC and BD, we get :

⇒ −2+a2\dfrac{-2 + a}{2} = 1

⇒ -2 + a = 2

⇒ a = 2 + 2

⇒ a = 4.

Equating the y-coordinates:

⇒ 1 = b2\dfrac{b}{2}

⇒ b = 2.

Hence, a = 4 and b = 2.

Question 22

The three vertices of a parallelogram ABCD, taken in order, are A(-1, 0), B(3, 1) and C(2, 2). Find the co-ordinates of the fourth vertex of the parallelogram.

Answer

Let coordinates of D be (x, y).

The three vertices of a parallelogram ABCD, taken in order, are A(-1, 0), B(3, 1) and C(2, 2). Find the co-ordinates of the fourth vertex of the parallelogram. Reflection, RSA Mathematics Solutions ICSE Class 10.

By using mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Mid-point of AC :

⇒MAC=(−1+22,0+22)⇒MAC=(12,1).\Rightarrow M_{AC} = \Big(\dfrac{-1 + 2}{2}, \dfrac{0 + 2}{2}\Big) \\[1em] \Rightarrow M_{AC} = \Big(\dfrac{1}{2}, 1\Big).

Midpoint of Diagonal BD :

⇒MBD=(3+x2,1+y2)\Rightarrow M_{BD} = \Big(\dfrac{3 + x}{2}, \dfrac{1 + y}{2}\Big)

Since its a parallelogram, the midpoint of diagonal AC must be equal to the midpoint of diagonal BD, as diagonals bisect each other.

Equating the x-coordinates:

⇒3+x2=12\Rightarrow \dfrac{3 + x}{2} = \dfrac{1}{2}

⇒ 3 + x = 1

⇒ x = 1 - 3

⇒ x = -2.

Equating the y-coordinates:

⇒1+y2=1\Rightarrow \dfrac{1 + y}{2} = 1

⇒ 1 + y = 2

⇒ y = 2 - 1

⇒ y = 1.

Hence, coordinates of D = (-2, 1).

Question 23

Find the lengths of the medians of a ΔABC whose vertices are A(-1, 3), B(1, -1) and C(5, 1). Also, find the co-ordinates of the centroid of ΔABC.

Answer

Using the Midpoint Formula,

(x,y)=(x1+x22,y1+y22)(x, y) = \Big( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \Big).

Let D be the midpoint of BC.

Find the lengths of the medians of a ΔABC whose vertices are A(-1, 3), B(1, -1) and C(5, 1). Also, find the co-ordinates of the centroid of ΔABC. Reflection, RSA Mathematics Solutions ICSE Class 10.

⇒D=(1+52,−1+12)⇒D=(62,02)⇒D=(3,0).\Rightarrow D = \Big( \dfrac{1 + 5}{2}, \dfrac{-1 + 1}{2} \Big) \\[1em] \Rightarrow D = \Big( \dfrac{6}{2}, \dfrac{0}{2} \Big) \\[1em] \Rightarrow D = (3, 0).

Let E be the midpoint of AC.

⇒E=(−1+52,3+12)⇒E=(42,42)⇒E=(2,2).\Rightarrow E = \Big( \dfrac{-1 + 5}{2}, \dfrac{3 + 1}{2} \Big) \\[1em] \Rightarrow E = \Big( \dfrac{4}{2}, \dfrac{4}{2} \Big) \\[1em] \Rightarrow E = (2, 2).

Let F be the midpoint of AB.

⇒F=(−1+12,3+(−1)2)⇒F=(02,22)⇒F=(0,1).\Rightarrow F = \Big( \dfrac{-1 + 1}{2}, \dfrac{3 + (-1)}{2} \Big) \\[1em] \Rightarrow F = \Big( \dfrac{0}{2}, \dfrac{2}{2} \Big) \\[1em] \Rightarrow F = (0, 1).

Length of Median AD : A(-1, 3) and D(3, 0)

Using Distance Formula,

D=(x2−x1)2+(y2−y1)2D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

Substituting values we get:

AD=(3−(−1))2+(0−3)2=(4)2+(−3)2=16+9=25=5 units.AD = \sqrt{(3 - (-1))^2 + (0 - 3)^2} \\[1em] = \sqrt{(4)^2 + (-3)^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.

Length of Median BE:

B(1, -1) and E(2, 2)

BE=(2−1)2+(2−(−1))2=(1)2+(3)2=1+9=10 units.BE = \sqrt{(2 - 1)^2 + (2 - (-1))^2} \\[1em]= \sqrt{(1)^2 + (3)^2} \\[1em] = \sqrt{1 + 9} \\[1em] = \sqrt{10} \text{ units}.

Length of Median CF:

C(5, 1) and F(0, 1)

CF=(0−5)2+(1−1)2=(−5)2+(0)2=25+0=25=5 units.CF = \sqrt{(0 - 5)^2 + (1 - 1)^2} \\[1em] = \sqrt{(-5)^2 + (0)^2} \\[1em] = \sqrt{25 + 0} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.

Centroid of triangle (G) =(x1+x2+x33,y1+y2+y33)= \Big( \dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3} \Big)

Substituting values we get :

⇒Centroid=(−1+1+53,3+(−1)+13)=(53,33)=(53,1).\Rightarrow \text{Centroid} = \Big(\dfrac{-1 + 1 + 5}{3}, \dfrac{3 + (-1) + 1}{3}\Big) \\[1em] = \Big(\dfrac{5}{3}, \dfrac{3}{3}\Big) \\[1em] = \Big(\dfrac{5}{3}, 1\Big).

Hence, AD = 5 units , BE=10BE = \sqrt{10} units, CF = 5 units, coordinates of centroid are (53,1)\Big(\dfrac{5}{3}, 1\Big).

Question 24

Find the co-ordinates of the centroid of ΔPQR whose vertices are P(6, 3), Q(-2, 5) and R(-1, 7).

Answer

By using the centroid formula,

(x,y)=(x1+x2+x33,y1+y2+y33)(x, y) = \Big( \dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3} \Big)

P(6, 3), Q(-2, 5) and R(-1, 7)

Find the co-ordinates of the centroid of ΔPQR whose vertices are P(6, 3), Q(-2, 5) and R(-1, 7). Reflection, RSA Mathematics Solutions ICSE Class 10.

Substitute values we get:

⇒(x,y)=(6+(−2)+(−1)3,3+5+73)⇒(x,y)=(6−33,153)⇒(x,y)=(33,153)⇒(x,y)=(1,5).\Rightarrow (x, y) = \Big( \dfrac{6 + (-2) + (-1)}{3}, \dfrac{3 + 5 + 7}{3} \Big) \\[1em] \Rightarrow (x, y) = \Big( \dfrac{6 - 3}{3}, \dfrac{15}{3} \Big) \\[1em] \Rightarrow (x, y) = \Big( \dfrac{3}{3}, \dfrac{15}{3} \Big) \\[1em] \Rightarrow (x, y) = (1, 5).

Hence, coordinates of centroid of ΔPQR = (1, 5).

Question 25

Find the co-ordinates of the point of intersection of the medians of the triangle whose vertices are A(-7, 5), B(-1, -3) and C(5, 7).

Answer

The medians of a triangle intersect at centroid of triangle.

By using the centroid formula,

(x,y)=(x1+x2+x33,y1+y2+y33)(x, y) = \Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3} \Big)

A(-7, 5), B(-1, -3) and C(5, 7)

Find the co-ordinates of the point of intersection of the medians of the triangle whose vertices are A(-7, 5), B(-1, -3) and C(5, 7). Reflection, RSA Mathematics Solutions ICSE Class 10.

Substitute values we get:

⇒(x,y)=(−7+(−1)+53,5+(−3)+73)⇒(x,y)=(−8+53,5+43)⇒(x,y)=(−33,93)⇒(x,y)=(−1,3).\Rightarrow (x, y) = \Big( \dfrac{-7 + (-1) + 5}{3}, \dfrac{5 + (-3) + 7}{3} \Big) \\[1em] \Rightarrow (x, y) = \Big( \dfrac{-8 + 5}{3}, \dfrac{5 + 4}{3} \Big) \\[1em] \Rightarrow (x, y) = \Big( \dfrac{-3}{3}, \dfrac{9}{3} \Big) \\[1em] \Rightarrow (x, y) = (-1, 3).

Hence, co-ordinates of the point of intersection of the medians = (-1, 3).

Question 26

If G(-2, 1) is the centroid of ΔABC, two of whose vertices are A(1, -6) and B(-5, 2), find the third vertex of the triangle.

Answer

Let coordinates of third vertex be C(x3, y3).

By using the centroid formula,

(x,y)=(x1+x2+x33,y1+y2+y33)(x, y) = \Big( \dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3} \Big)

Given,

G(-2, 1) is the centroid.

If G(-2, 1) is the centroid of ΔABC, two of whose vertices are A(1, -6) and B(-5, 2), find the third vertex of the triangle. Reflection, RSA Mathematics Solutions ICSE Class 10.

Substituting values we get :

⇒(−2,1)=(1+(−5)+x33,−6+2+y33)⇒−2=(1+(−5)+x33),1=(−6+2+y33)⇒−6=1+(−5)+x3,3=−6+2+y3⇒−6=−4+x3,3=−4+y3⇒−6+4=x3,3+4=y3⇒x3=−2,y3=7.\Rightarrow (-2, 1) = \Big( \dfrac{1 + (-5) + x_3}{3}, \dfrac{-6 + 2 + y_3}{3}\Big) \\[1em] \Rightarrow -2 = \Big( \dfrac{1 + (-5) + x_3}{3}\Big), 1 = \Big( \dfrac{-6 + 2 + y_3}{3}\Big) \\[1em] \Rightarrow -6 = 1 + (-5) + x_3, 3 = -6 + 2 + y_3 \\[1em] \Rightarrow -6 = -4 + x_3, 3 = -4 + y_3 \\[1em] \Rightarrow -6 + 4 = x_3, 3 + 4 = y_3 \\[1em] \Rightarrow x_3 = -2, y_3 = 7.

Hence, coordinates of third vertex = (-2, 7).

Question 27

A(6, y), B(-4, 4) and C(x, -1) are the vertices of ΔABC whose centroid is the origin. Calculate the values of x and y.

Answer

By using the centroid formula,

(x,y)=(x1+x2+x33,y1+y2+y33)(x, y) = \Big( \dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3} \Big)

Origin (0,0) is the centroid of triangle ABC.

A(6, y), B(-4, 4) and C(x, -1) are the vertices of ΔABC whose centroid is the origin. Calculate the values of x and y. Reflection, RSA Mathematics Solutions ICSE Class 10.

Solving for x-coordinate:

⇒0=(6+(−4)+x3)⇒0=2+x⇒x=−2.\Rightarrow 0 = \Big(\dfrac{6 + (-4) + x}{3}\Big) \\[1em] \Rightarrow 0 = 2 + x \\[1em] \Rightarrow x = -2.

Solving for y-coordinate:

⇒0=(y+4+(−1)3)⇒0=y+3⇒y=−3.\Rightarrow 0 = \Big(\dfrac{y + 4 + (-1)}{3}\Big) \\[1em] \Rightarrow 0 = y + 3 \\[1em] \Rightarrow y = -3.

Hence, x = -2 and y = -3.

Question 28

ABC is a triangle and G(4, 3) is its centroid. If A(1, 3), B(4, b) and C(a, 1) be the vertices, find the values of a and b and hence find the length of side BC.

Answer

By using the centroid formula,

(x,y)=(x1+x2+x33,y1+y2+y33)(x, y) = \Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)

Given,

A(1, 3), B(4, b) and C(a, 1).

G(4, 3) is the centroid of triangle ABC.

ABC is a triangle and G(4, 3) is its centroid. If A(1, 3), B(4, b) and C(a, 1) be the vertices, find the values of a and b and hence find the length of side BC. Reflection, RSA Mathematics Solutions ICSE Class 10.

Solving for x-coordinate :

⇒4=1+4+a3⇒12=5+a⇒a=12−5⇒a=7.\Rightarrow 4 = \dfrac{1 + 4 + a}{3} \\[1em] \Rightarrow 12 = 5 + a \\[1em] \Rightarrow a = 12 - 5 \\[1em] \Rightarrow a = 7.

C = (a, 1) = (7, 1)

Solving for y-coordinate :

⇒3=3+b+13⇒9=4+b⇒b=9−4⇒b=5.\Rightarrow 3 = \dfrac{3 + b + 1}{3} \\[1em] \Rightarrow 9 = 4 + b \\[1em] \Rightarrow b = 9 - 4 \\[1em] \Rightarrow b = 5.

B = (4, b) = (4, 5)

By using distance formula,

D=(x2−x1)2+(y2−y1)2D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

Substituting values we get:

BC=(7−4)2+(1−5)2=(3)2+(−4)2=9+16=25=5 units.BC = \sqrt{(7 - 4)^2 + (1 - 5)^2} \\[1em] = \sqrt{(3)^2 + (-4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.

Hence, a = 7 and b = 5, length of BC = 5 units.

Question 29

Calculate the ratio in which the line segment joining A(-4, 2) and B(3, 6) is divided by the point P(x, 3). Also, find

(i) x

(ii) length of AP.

Answer

(i) Let the point P(x, 3) divide the line segment joining A(-4, 2) and B(3, 6) in the ratio m1 : m2.

Calculate the ratio in which the line segment joining A(-4, 2) and B(3, 6) is divided by the point P(x, 3). Also, find Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒3=(m1(6)+m2(2)m1+m2)⇒3(m1+m2)=6m1+2m2⇒3m1+3m2=6m1+2m2⇒3m2−2m2=6m1−3m1⇒m2=3m1⇒m1m2=13.\Rightarrow 3 = \Big(\dfrac{m_1(6) + m_2(2)}{m_1 + m_2}\Big) \\[1em] \Rightarrow 3(m_1 + m_2) = {6m_1 + 2m_2} \\[1em] \Rightarrow 3m_1 + 3m_2 = 6m_1 + 2m_2 \\[1em] \Rightarrow 3m_2 - 2m_2 = 6m_1 - 3m_1 \\[1em] \Rightarrow m_2 = 3m_1 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{1}{3}.

Solving for x-coordinate with the ratio m1=1m_1 = 1 and m2=3m_2 = 3:

⇒x=(1)(3)+(3)(−4)1+3=3−124=−94.\Rightarrow x = \dfrac{(1)(3) + (3)(-4)}{1 + 3} \\[1em] = \frac{3 - 12}{4} \\[1em] = \frac{-9}{4}.

Hence, x = −94\dfrac{-9}{4}.

(ii) Solving length of AP

By using Distance Formula

D=(x2−x1)2+(y2−y1)2D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

A(-4, 2) and P(−94,3)P\Big(\dfrac{-9}{4}, 3\Big).

AP=(−94−(−4))2+(3−2)2=(−94+164)2+(1)2=(74)2+12=4916+1616=6516=654 unitsAP = \sqrt{\Big(-\dfrac{9}{4} - (-4)\Big)^2 + (3 - 2)^2} \\[1em] =\sqrt{\Big(-\dfrac{9}{4} + \dfrac{16}{4}\Big)^2 + (1)^2} \\[1em] = \sqrt{\Big(\dfrac{7}{4}\Big)^2 + 1^2} \\[1em] = \sqrt{\dfrac{49}{16} + \dfrac{16}{16}} \\[1em] = \sqrt{\dfrac{65}{16}} \\[1em] = \dfrac{\sqrt{65}}{4} \text{ units}

Hence, length of AP = 654\dfrac{\sqrt{65}}{4} units.

Question 30

In what ratio is the line joining P(5, 3) and Q(-5, 3) divided by the x-axis? Also, find the co-ordinates of the point of intersection.

Answer

Let point on x-axis dividing PQ be (a, 0).

By section formula,

y = m1y2+m2y1m1+m2\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}

Substituting values we get :

⇒0=m1×3+m2×3m1+m2⇒0=3m1+3m2⇒3m1=−3m2⇒m1m2=−33⇒m1m2=−11\Rightarrow 0 = \dfrac{m_1 \times 3 + m_2 \times 3}{m_1 + m_2} \\[1em] \Rightarrow 0 = 3m_1 + 3m_2 \\[1em] \Rightarrow 3m_1 = -3m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = -\dfrac{3}{3} \\[1em] \Rightarrow \dfrac{m_1}{m_2} = -\dfrac{1}{1}

The negative ratio shows that the line PQ does not intersects with x-axis.

Hence, the x-axis does not intersect the line PQ, so no ratio or point of intersection exists.

Question 31

Find a point P which divides internally the line segment joining the points A(-3, 9) and B(1, -3) in the ratio 1 : 3.

Answer

Let Point P divides the line segment joining A(−3, 9) and B(1, −3) internally in the ratio m1 : m2 = 1 : 3.

Find a point P which divides internally the line segment joining the points A(-3, 9) and B(1, -3) in the ratio 1 : 3. Reflection, RSA Mathematics Solutions ICSE Class 10.

By using section formula,

P(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get:

P=(1×1+3×−31+3,1×−3+3×91+3)=(1−94,−3+274)=(−84,244)=(−2,6).P = \Big(\dfrac{1 \times 1 + 3 \times -3}{1 + 3}, \dfrac{1 \times -3 + 3 \times 9}{1 + 3}\Big) \\[1em] = \Big(\dfrac{1 - 9}{4}, \dfrac{-3 + 27}{4}\Big) \\[1em] = \Big(\dfrac{-8}{4}, \dfrac{24}{4}\Big) \\[1em] = (-2, 6).

Hence, coordinates of P = (-2, 6).

Multiple Choice Questions

Question 1

The coordinates of the point P which divides the join of A(5, -2) and B(9, 6) in the ratio 3 : 1 are :

  1. (4, -7)

  2. (72,4)\Big(\dfrac{7}{2}, 4\Big)

  3. (8, 4)

  4. (12, 8)

Answer

Let point P be (x, y).

The coordinates of the point P which divides the join of A(5, -2) and B(9, 6) in the ratio 3 : 1 are : Reflection, RSA Mathematics Solutions ICSE Class 10.

Given,

m1 : m2 = 3 : 1

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒(x,y)=(3×9+1×53+1,3×6+1×−23+1)=(27+54,18−24)=(324,164)=(8,4).\Rightarrow (x, y) = \Big(\dfrac{3 \times 9 + 1 \times 5}{3 + 1}, \dfrac{3 \times 6 + 1 \times -2}{3 + 1}\Big) \\[1em] = \Big(\dfrac{27 + 5}{4}, \dfrac{18 - 2}{4}\Big) \\[1em] = \Big(\dfrac{32}{4}, \dfrac{16}{4}\Big) \\[1em] = (8, 4).

Hence, Option 3 is the correct option.

Question 2

The coordinates of the point on x-axis which divides the line segment joining the points (2, 3) and (5, -6) in the ratio 1 : 2 are :

  1. (2, 0)

  2. (-2, 0)

  3. (3, 0)

  4. (-3, 0)

Answer

Let point P be (x, y).

The coordinates of the point on x-axis which divides the line segment joining the points (2, 3) and (5, -6) in the ratio 1 : 2 are : Reflection, RSA Mathematics Solutions ICSE Class 10.

Given,

m1 : m2 = 1 : 2

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒(x,y)=(1×5+2×21+2,1×(−6)+2×31+2)=(5+43,−6+63)=(93,03)=(3,0).\Rightarrow (x, y) = \Big(\dfrac{1 \times 5 + 2 \times 2}{1 + 2}, \dfrac{1 \times (-6) + 2 \times 3}{1 + 2}\Big) \\[1em] = \Big(\dfrac{5 + 4}{3}, \dfrac{-6 + 6}{3}\Big) \\[1em] = \Big(\dfrac{9}{3}, \dfrac{0}{3}\Big) \\[1em] = (3, 0).

Hence, Option 3 is the correct option.

Question 3

The point which divides the line segment joining the points A(3, -2) and B(6, 7) internally in the ratio 3 : 2 lies in which of the following quadrants?

  1. I

  2. II

  3. III

  4. IV

Answer

Let point P be (x, y).

The point which divides the line segment joining the points A(3, -2) and B(6, 7) internally in the ratio 3 : 2 lies in which of the following quadrants? Reflection, RSA Mathematics Solutions ICSE Class 10.

Given,

m1 : m2 = 3 : 2

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒(x,y)=(3×6+2×33+2,3×7+2×(−2)3+2)=(18+65,21−45)=(245,175).\Rightarrow (x, y) = \Big(\dfrac{3 \times 6 + 2 \times 3}{3 + 2}, \dfrac{3 \times 7 + 2 \times (-2)}{3 + 2}\Big) \\[1em] = \Big(\dfrac{18 + 6}{5}, \dfrac{21 - 4}{5}\Big) \\[1em] = \Big(\dfrac{24}{5}, \dfrac{17}{5}\Big).

Here, both x and y are positive.

Therefore, the point lies in the 1st Quadrant.

Hence, Option 1 is the correct option.

Question 4

If the point R(k, 4) divides the line segment joining the points P(2, 6) and Q(5, 1) in the ratio 2 : 3, then the value of k is:

  1. -5

  2. (−165)\Big(\dfrac{-16}{5}\Big)

  3. 5

  4. (165)\Big(\dfrac{16}{5}\Big)

Answer

Given,

R = (k, 4)

m1 : m2 = 2 : 3

R(k, 4) divides the line segment joining the points P(2, 6) and Q(5, 1) in the ratio 2 : 3.

If the point R(k, 4) divides the line segment joining the points P(2, 6) and Q(5, 1) in the ratio 2 : 3, then the value of k is: Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒(k,4)=(2×5+3×22+3,2×1+3×62+3)⇒(k,4)=(10+65,2+185)⇒(k,4)=(165,205)⇒(k,4)=(165,4).\Rightarrow (k, 4) = \Big(\dfrac{2 \times 5 + 3 \times 2}{2 + 3}, \dfrac{2 \times 1 + 3 \times 6}{2 + 3}\Big) \\[1em] \Rightarrow (k, 4) = \Big(\dfrac{10 + 6}{5}, \dfrac{2 + 18}{5}\Big) \\[1em] \Rightarrow (k, 4) = \Big(\dfrac{16}{5}, \dfrac{20}{5}\Big) \\[1em] \Rightarrow (k, 4) = \Big(\dfrac{16}{5}, 4\Big).

Thus, k = 165\dfrac{16}{5}.

Hence, Option 4 is the correct option.

Question 5

If the point P(k, 0) divides the line segment joining the points A(2, -2) and B(-7, 4) in the ratio 1 : 2, then the value of k is:

  1. -2

  2. -1

  3. 1

  4. 2

Answer

Given,

P = (k, 0)

m1 : m2 = 1 : 2

P(k, 0) divides the line segment joining the points A(2, -2) and B(-7, 4) in the ratio 1 : 2.

If the point P(k, 0) divides the line segment joining the points A(2, -2) and B(-7, 4) in the ratio 1 : 2, then the value of k is: Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒(k,0)=(1×(−7)+2×21+2,1×4+2×(−2)1+2)⇒(k,0)=(−7+43,4−43)⇒(k,0)=(−33,03)⇒(k,0)=(−1,0).\Rightarrow (k, 0) = \Big(\dfrac{1 \times (-7) + 2 \times 2}{1 + 2}, \dfrac{1 \times 4 + 2 \times (-2)}{1 + 2}\Big) \\[1em] \Rightarrow (k, 0) = \Big(\dfrac{-7 + 4}{3}, \dfrac{4 - 4}{3}\Big) \\[1em] \Rightarrow (k, 0) = \Big(\dfrac{-3}{3}, \dfrac{0}{3}\Big) \\[1em] \Rightarrow (k, 0) = (-1, 0).

Thus, k = -1.

Hence, Option 2 is the correct option.

Question 6

If the point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3 : 1, then the value of y is :

  1. 1

  2. 2

  3. 3

  4. 4

Answer

Let point P be (6, 2).

If the point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3 : 1, then the value of y is : Reflection, RSA Mathematics Solutions ICSE Class 10.

Given,

m1 : m2 = 3 : 1

P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3 : 1.

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

We use the y–coordinate to find y.

⇒2=3×y+1×53+1⇒2=3y+54⇒8=3y+5⇒3y=3⇒y=1.\Rightarrow 2 = \dfrac{3 \times y + 1 \times 5}{3 + 1} \\[1em] \Rightarrow 2 = \dfrac{3y + 5}{4} \\[1em] \Rightarrow 8 = 3y + 5 \\[1em] \Rightarrow 3y = 3 \\[1em] \Rightarrow y = 1.

Hence, Option 1 is the correct option.

Question 7

The ratio in which the point P(1, 2) divides the join of the points A(-2, 1) and B(7, 4) is:

  1. 1 : 2

  2. 2 : 1

  3. 3 : 2

  4. 2 : 3

Answer

Let the ratio in which P divides AB be k : 1.

The ratio in which the point P(1, 2) divides the join of the points A(-2, 1) and B(7, 4) is: Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Using the y–coordinate to find ratio.

⇒2=(k(4)+1(1)k+1)⇒2=(4k+1k+1)⇒2(k+1)=4k+1⇒2k+2=4k+1⇒2−1=4k−2k⇒2k=1⇒k=12⇒k:1=12:1=1:2.\Rightarrow 2 = \Big(\dfrac{k(4) + 1(1)}{k + 1}\Big) \\[1em] \Rightarrow 2 = \Big(\dfrac{4k + 1}{k + 1}\Big) \\[1em] \Rightarrow 2(k + 1) = 4k + 1 \\[1em] \Rightarrow 2k + 2 = 4k + 1 \\[1em] \Rightarrow 2 - 1 = 4k - 2k \\[1em] \Rightarrow 2k = 1 \\[1em] \Rightarrow k = \dfrac{1}{2}\\[1em] \Rightarrow k : 1 = \dfrac{1}{2} : 1 = 1 : 2.

Hence, Option 1 is the correct option.

Question 8

The ratio in which the line segment joining A(2, -3) and B(5, 6) is divided by x-axis is :

  1. 1 : 2

  2. 2 : 1

  3. 3 : 5

  4. 2 : 3

Answer

Let the point where x-axis intersects the line segment be P(x, 0).

Let the ratio in which P divides AB be m1 : m2.

The ratio in which the line segment joining A(2, -3) and B(6, 5) is divided by x-axis is : Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula (using y–coordinate),

⇒y=(m1y2+m2y1m1+m2)⇒0=m1×5+m2×(−3)m1+m2⇒0=5m1−3m2⇒5m1=3m2⇒m1m2=35.\Rightarrow y = \Big(\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] \Rightarrow 0 = \dfrac{m_1 \times 5 + m_2 \times (-3)}{m_1 + m_2} \\[1em] \Rightarrow 0 = 5m_1 - 3m_2 \\[1em] \Rightarrow 5m_1 = 3m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{3}{5}.

Thus, the required ratio is 3 : 5.

Hence, Option 3 is the correct option.

Question 9

In what ratio is the line segment joining the points P(-4, 2) and Q(8, 3) divided by y-axis?

  1. 1 : 3

  2. 3 : 1

  3. 1 : 2

  4. 2 : 1

Answer

Let the point where y-axis divides the line segment be R(0, y).

Let the ratio be m1 : m2.

In what ratio is the line segment joining the points P(-4, 2) and Q(8, 3) divided by y-axis? Reflection, RSA Mathematics Solutions ICSE Class 10.

Using section-formula,

⇒x=m1x2+m2x1m1+m2⇒0=m1×8+m2×(−4)m1+m2⇒0=8m1−4m2⇒8m1=4m2⇒m1m2=48=12.\Rightarrow x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 0 = \dfrac{m_1 \times 8 + m_2 \times (-4)}{m_1 + m_2} \\[1em] \Rightarrow 0 = 8m_1 - 4m_2 \\[1em] \Rightarrow 8m_1 = 4m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{4}{8} = \dfrac{1}{2}.

Thus, the required ratio is 1 : 2.

Hence, Option 3 is the correct option.

Question 10

Point P divides the line segment joining R(-1, 3) and S(9, 8) in the ratio k : 1. If P lies on the line x - y + 2 = 0, then the value of k is:

  1. 12\dfrac{1}{2}

  2. 13\dfrac{1}{3}

  3. 14\dfrac{1}{4}

  4. 23\dfrac{2}{3}

Answer

Let point P be (x, y).

Point P divides the line segment joining R(-1, 3) and S(9, 8) in the ratio k : 1. If P lies on the line x - y + 2 = 0, then the value of k is: Reflection, RSA Mathematics Solutions ICSE Class 10.

Given,

m1 : m2 = k : 1

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒(x,y)=(k×9+1×(−1)k+1,k×8+1×3k+1)=(9k−1k+1,8k+3k+1).\Rightarrow (x, y) = \Big(\dfrac{k \times 9 + 1 \times (-1)}{k + 1}, \dfrac{k \times 8 + 1 \times 3}{k + 1}\Big) \\[1em] = \Big(\dfrac{9k - 1}{k + 1}, \dfrac{8k + 3}{k + 1}\Big).

Since P lies on the line x - y + 2 = 0, substituting values of x and y:

⇒9k−1k+1−8k+3k+1+2=0⇒9k−1−(8k+3)k+1+2=0⇒9k−1−8k−3k+1+2=0⇒k−4k+1+2=0⇒k−4+2(k+1)k+1=0⇒k−4+2k+2k+1=0⇒3k−2k+1=0⇒3k−2=0⇒3k=2⇒k=23.\Rightarrow \dfrac{9k - 1}{k + 1} - \dfrac{8k + 3}{k + 1} + 2 = 0 \\[1em] \Rightarrow \dfrac{9k - 1 - (8k + 3)}{k + 1} + 2 = 0 \\[1em] \Rightarrow \dfrac{9k - 1 - 8k - 3}{k + 1} + 2 = 0 \\[1em] \Rightarrow \dfrac{k - 4}{k + 1} + 2 = 0 \\[1em] \Rightarrow \dfrac{k - 4 + 2(k + 1)}{k + 1} = 0 \\[1em] \Rightarrow \dfrac{k - 4 + 2k + 2}{k + 1} = 0 \\[1em] \Rightarrow \dfrac{3k - 2}{k + 1} = 0 \\[1em] \Rightarrow 3k - 2 = 0 \\[1em] \Rightarrow 3k = 2 \\[1em] \Rightarrow k = \dfrac{2}{3}.

Hence, Option 4 is the correct option.

Question 11

In the adjoining figure, P(5, -3) and Q(3, y) are the points of trisection of the line segment joining A(7, -2) and B(1, -5). Then, y equals :

In the adjoining figure, P(5, -3) and Q(3, y) are the points of trisection of the line segment joining A(7, -2) and B(1, -5). Then, y equals : Reflection, RSA Mathematics Solutions ICSE Class 10.

  1. -4

  2. (−52)\Big(\dfrac{-5}{2}\Big)

  3. 2

  4. 4

Answer

Since P and Q trisect the line segment AB, the point Q(3, y) divides A(7, -2) and B(1, -5) in the ratio 2 : 1.

Let point Q be (3, y).

Given,

m1 : m2 = 2 : 1

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒(3,y)=(2×1+1×72+1,2×(−5)+1×(−2)2+1)⇒(3,y)=(2+73,−10−23)⇒(3,y)=(93,−123)⇒(3,y)=(3,−4).\Rightarrow (3, y) = \Big(\dfrac{2 \times 1 + 1 \times 7}{2 + 1}, \dfrac{2 \times (-5) + 1 \times (-2)}{2 + 1}\Big) \\[1em] \Rightarrow (3, y) = \Big(\dfrac{2 + 7}{3}, \dfrac{-10 - 2}{3}\Big) \\[1em] \Rightarrow (3, y) = \Big(\dfrac{9}{3}, \dfrac{-12}{3}\Big) \\[1em] \Rightarrow (3, y) = (3, -4).

Thus, y = -4.

Hence, Option 1 is the correct option.

Question 12

If the point P(6, -3) lies on the line segment joining points A(4, 2) and B(8, 4), then:

  1. AP = 34\dfrac{3}{4} AB

  2. AP = 14\dfrac{1}{4} AB

  3. PB = 13\dfrac{1}{3} AB

  4. AP = 12\dfrac{1}{2} AB

Answer

Let the ratio in which P divides AB be k : 1.

Draw co-ordinate axes and represent the following points : Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula,

x = m1x2+m2x1m1+m2\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}

Substituting values we get :

⇒6=(k(8)+1(4)k+1)⇒6=(8k+4k+1)⇒6(k+1)=8k+4⇒6k+6=8k+4⇒6−4=8k−6k⇒2=2k⇒k=11=1:1.\Rightarrow 6 = \Big(\dfrac{k(8) + 1(4)}{k + 1}\Big) \\[1em] \Rightarrow 6 = \Big(\dfrac{8k + 4}{k + 1}\Big) \\[1em] \Rightarrow 6(k + 1) = 8k + 4 \\[1em] \Rightarrow 6k + 6 = 8k + 4 \\[1em] \Rightarrow 6 - 4 = 8k - 6k \\[1em] \Rightarrow 2 = 2k \\[1em] \Rightarrow k = \dfrac{1}{1} = 1:1.

This means that P is the midpoint of AB.

∴ AP = 12\dfrac{1}{2} AB.

Hence, Option 4 is the correct option.

Question 13

The mid-point of the line segment joining the points (-3, 2) and (7, 6) is:

  1. (-2, -4)

  2. (-2, 4)

  3. (2, 4)

  4. (4, 2)

Answer

Let the mid-point be M(x, y).

The mid-point of the line segment joining the points (-3, 2) and (7, 6) is: Reflection, RSA Mathematics Solutions ICSE Class 10.

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Substituting values we get :

⇒M(x,y)=(−3+72,2+62)⇒(42,82)⇒(2,4).\Rightarrow M(x, y) = \Big(\dfrac{-3 + 7}{2}, \dfrac{2 + 6}{2}\Big) \\[1em] \Rightarrow \Big(\dfrac{4}{2}, \dfrac{8}{2}\Big) \\[1em] \Rightarrow (2, 4).

Hence, Option 3 is the correct option.

Question 14

If A(4, 2), B(6, 5) and C(1, 4) be the vertices of ΔABC and AD is a median, then the coordinates of D are:

  1. (52,3)\Big(\dfrac{5}{2}, 3\Big)

  2. (5,72)\Big(5, \dfrac{7}{2}\Big)

  3. (72,92)\Big(\dfrac{7}{2}, \dfrac{9}{2}\Big)

  4. none of these

Answer

Since AD is a median, D is the mid-point of BC.

Let point D be (x, y).

If A(4, 2), B(6, 5) and C(1, 4) be the vertices of ΔABC and AD is a median, then the coordinates of D are: Reflection, RSA Mathematics Solutions ICSE Class 10.

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

⇒(x,y)=(6+12,5+42)⇒(72,92).\Rightarrow (x, y) = \Big(\dfrac{6 + 1}{2}, \dfrac{5 + 4}{2}\Big) \\[1em] \Rightarrow \Big(\dfrac{7}{2}, \dfrac{9}{2}\Big).

Hence, Option 3 is the correct option.

Question 15

If (3, -6) is the mid-point of the line segment joining (0, 0) and (x, y), then the point (x, y) is:

  1. (-3, 6)

  2. (6, -6)

  3. (6, -12)

  4. (32,−3)\Big(\dfrac{3}{2}, -3\Big)

Answer

Given, (3, -6) is the mid-point of the line segment joining (0, 0) and (x, y).

If (3, -6) is the mid-point of the line segment joining (0, 0) and (x, y), then the point (x, y) is: Reflection, RSA Mathematics Solutions ICSE Class 10.

∴(3,−6)=(0+x2,0+y2)⇒(3,−6)=(x2,y2)⇒3=x2 and −6=y2⇒x=6 and y=−12\therefore (3, -6) = \Big(\dfrac{0 + x}{2}, \dfrac{0 + y}{2}\Big) \\[1em] \Rightarrow (3, -6) = \Big(\dfrac{x}{2}, \dfrac{y}{2}\Big) \\[1em] \Rightarrow 3 = \dfrac{x}{2} \text{ and } -6 = \dfrac{y}{2} \\[1em] \Rightarrow x = 6 \text{ and } y = -12

(x, y) = (6, -12).

Hence, Option 3 is the correct option.

Question 16

Point P(a8,4)P\Big(\dfrac{a}{8}, 4\Big) is the mid-point of the line segment joining the points A(-5, 2) and B(4, 6). Then, the value of a is:

  1. -8

  2. -4

  3. 2

  4. 4

Answer

Given,

P(a8,4)P\Big(\dfrac{a}{8}, 4\Big) is the mid-point of the line segment joining the points A(-5, 2) and B(4, 6).

id-point of the line segment joining the points A(-5, 2) and B(4, 6). Then, the value of a is:Reflection, RSA Mathematics Solutions ICSE Class 10.

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Substituting values we get :

⇒(a8,4)=(−5+42,2+62)⇒(a8,4)=(−12,4)\Rightarrow \Big(\dfrac{a}{8}, 4\Big) = \Big(\dfrac{-5 + 4}{2}, \dfrac{2 + 6}{2}\Big) \\[1em] \Rightarrow \Big(\dfrac{a}{8}, 4\Big) = \Big(\dfrac{-1}{2}, 4\Big)

Comparing the x-coordinates, we get :

⇒a8=−12⇒a=−82⇒a=−4.\Rightarrow \dfrac{a}{8} = \dfrac{-1}{2} \\[1em] \Rightarrow a = \dfrac{-8}{2} \\[1em] \Rightarrow a = -4.

Hence, Option 2 is the correct option.

Question 17

A(-3, b) and B(1, b + 4) are two points. If the coordinates of the mid-point of AB are (-1, 1), then the value of b is :

  1. -1

  2. 0

  3. 1

  4. 2

Answer

Given,

Mid-point of AB = (-1, 1).

A(-3, b) and B(1, b + 4) are two points. If the coordinates of the mid-point of AB are (-1, 1), then the value of b is : Reflection, RSA Mathematics Solutions ICSE Class 10.

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Substituting values we get :

⇒(−1,1)=(−3+12,b+(b+4)2)⇒(−1,1)=(−22,2b+42)⇒(−1,1)=(−1,b+2)\Rightarrow (-1, 1) = \Big(\dfrac{-3 + 1}{2}, \dfrac{b + (b + 4)}{2}\Big) \\[1em] \Rightarrow (-1, 1) = \Big(\dfrac{-2}{2}, \dfrac{2b + 4}{2}\Big) \\[1em] \Rightarrow (-1, 1) = (-1, b + 2)

Comparing the y-coordinates, we get :

⇒ 1 = b + 2

⇒ b = -1.

Hence, Option 1 is the correct option.

Question 18

If the point R(5, 7) is the mid-point of the line segment joining the points P(3, y) and Q(x, 9), then (x + y) equals:

  1. 7

  2. 9

  3. 12

  4. 14

Answer

Given,

R(5, 7) is the mid-point of the line segment joining the points P(3, y) and Q(x, 9).

If the point R(5, 7) is the mid-point of the line segment joining the points P(3, y) and Q(x, 9), then (x + y) equals: Reflection, RSA Mathematics Solutions ICSE Class 10.

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Substituting values we get :

⇒(5,7)=(3+x2,y+92)⇒5=3+x2 and 7=y+92⇒10=3+x and 14=y+9⇒x=7 and y=5.\Rightarrow (5, 7) = \Big(\dfrac{3 + x}{2}, \dfrac{y + 9}{2}\Big) \\[1em] \Rightarrow 5 = \dfrac{3 + x}{2} \text{ and } 7 = \dfrac{y + 9}{2} \\[1em] \Rightarrow 10 = 3 + x \text{ and } 14 = y + 9 \\[1em] \Rightarrow x = 7 \text{ and } y = 5.

x + y = 7 + 5 = 12.

Hence, Option 3 is the correct option.

Question 19

If the point (2, 1) is the mid-point of the line segment PQ joining the points P(92,−4)P\Big(\dfrac{9}{2}, -4\Big) and Q(a, b), then (a + b) is equal to:

  1. 74\dfrac{7}{4}

  2. 114\dfrac{11}{4}

  3. 72\dfrac{7}{2}

  4. 112\dfrac{11}{2}

Answer

Let the mid-point of PQ be (2, 1).

If the point (2, 1) is the mid-point of the line segment PQ joining the points Reflection, RSA Mathematics Solutions ICSE Class 10.

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Substituting values we get :

⇒(2,1)=(92+a2,−4+b2)\Rightarrow (2, 1) = \Big(\dfrac{\dfrac{9}{2} + a}{2}, \dfrac{-4 + b}{2}\Big)

Comparing the x coordinates, we get :

⇒92+a2=2⇒92+a=4⇒a=4−92⇒a=8−92=−12.\Rightarrow \dfrac{\dfrac{9}{2} + a}{2} = 2 \\[1em] \Rightarrow \dfrac{9}{2} + a = 4 \\[1em] \Rightarrow a = 4 - \dfrac{9}{2} \\[1em] \Rightarrow a = \dfrac{8 - 9}{2} = \dfrac{-1}{2}.

Comparing the y coordinates, we get :

⇒ −4+b2=1\dfrac{-4 + b}{2} = 1

⇒ -4 + b = 2

⇒ b = 6.

⇒a+b⇒−12+6⇒−1+122⇒112.\Rightarrow a + b \\[1em] \Rightarrow \dfrac{-1}{2} + 6 \\[1em] \Rightarrow \dfrac{-1 + 12}{2} \\[1em] \Rightarrow \dfrac{11}{2}.

Hence, Option 4 is the correct option.

Question 20

The coordinates of the vertices of ΔABC are respectively (-4, -2), (6, 2) and (4, 6). The centroid G of ΔABC is:

  1. (2, 2)

  2. (2, 3)

  3. (3, 3)

  4. (0, -1)

Answer

The coordinates of the vertices of ΔABC are respectively (-4, -2), (6, 2) and (4, 6). The centroid G of ΔABC is: Reflection, RSA Mathematics Solutions ICSE Class 10.

By centroid formula,

G(x, y) = (x1+x2+x33,y1+y2+y33)\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)

Substituting values we get :

⇒G=(−4+6+43,−2+2+63)=(63,63)=(2,2).\Rightarrow G = \Big(\dfrac{-4 + 6 + 4}{3}, \dfrac{-2 + 2 + 6}{3}\Big) \\[1em] = \Big(\dfrac{6}{3}, \dfrac{6}{3}\Big) \\[1em] = (2, 2).

Hence, Option 1 is the correct option.

Question 21

Two vertices of a ΔABC are A(-1, 4) and B(5, 2) and its centroid is (0, -3). The coordinates of C are :

  1. (4, 3)

  2. (4, 15)

  3. (-4, -15)

  4. (-15, -4)

Answer

Let the coordinates of C be (x, y).

Two vertices of a ΔABC are A(-1, 4) and B(5, 2) and its centroid is (0, -3). The coordinates of C are : Reflection, RSA Mathematics Solutions ICSE Class 10.

By centroid formula,

Centroid = (x1+x2+x33,y1+y2+y33)\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)

Given,

G(0, -3) is the centroid of the triangle.

Substituting values we get :

⇒(0,−3)=(−1+5+x3,4+2+y3)⇒(0,−3)=(4+x3,6+y3)⇒0=4+x3 and −3=6+y3⇒4+x=0 and 6+y=−9⇒x=−4 and y=−9−6⇒x=−4 and y=−15.\Rightarrow (0, -3) = \Big(\dfrac{-1 + 5 + x}{3}, \dfrac{4 + 2 + y}{3}\Big) \\[1em] \Rightarrow (0, -3) = \Big(\dfrac{4 + x}{3}, \dfrac{6 + y}{3}\Big) \\[1em] \Rightarrow 0 = \dfrac{4 + x}{3} \text{ and } -3 = \dfrac{6 + y}{3} \\[1em] \Rightarrow 4 + x = 0 \text{ and } 6 + y = -9 \\[1em] \Rightarrow x = -4 \text{ and } y = -9 - 6 \\[1em] \Rightarrow x = -4 \text{ and } y = -15.

C = (x, y) = (-4, -15).

Hence, Option 3 is the correct option.

Question 22

The mid-point of the line segment joining (4p, 5) and (2, 3q) is (5, 5p - 1). The values of p and q are respectively:

  1. 2, 83\dfrac{8}{3}

  2. -2, 83\dfrac{8}{3}

  3. 2, 133\dfrac{13}{3}

  4. -2, 133\dfrac{13}{3}

Answer

Given,

Mid-point of the line segment joining (4p, 5) and (2, 3q) is (5, 5p - 1).

The mid-point of the line segment joining (4p, 5) and (2, 3q) is (5, 5p - 1). The values of p and q are respectively: Reflection, RSA Mathematics Solutions ICSE Class 10.

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Substituting values we get :

⇒(5,5p−1)=(4p+22,5+3q2)⇒(5,5p−1)=(2(2p+1)2,5+3q2)⇒(5,5p−1)=(2p+1,5+3q2)\Rightarrow (5, 5p - 1) = \Big(\dfrac{4p + 2}{2}, \dfrac{5 + 3q}{2}\Big) \\[1em] \Rightarrow (5, 5p - 1) = \Big(\dfrac{2(2p + 1)}{2}, \dfrac{5 + 3q}{2}\Big) \\[1em] \Rightarrow (5, 5p - 1) = \Big(2p + 1, \dfrac{5 + 3q}{2}\Big)

Comparing the x coordinates, we get :

⇒ 2p + 1 = 5

⇒ 2p = 5 - 1

⇒ 2p = 4

⇒ p = 42\dfrac{4}{2}

⇒ p = 2.

Comparing y-coordinates we get :

⇒5p−1=5+3q2⇒5×2−1=5+3q2⇒9=5+3q2⇒18=5+3q⇒3q=18−5⇒3q=13⇒q=133.\Rightarrow 5p - 1 = \dfrac{5 + 3q}{2} \\[1em] \Rightarrow 5 \times 2 - 1 = \dfrac{5 + 3q}{2} \\[1em] \Rightarrow 9 = \dfrac{5 + 3q}{2} \\[1em] \Rightarrow 18 = 5 + 3q \\[1em] \Rightarrow 3q = 18 - 5 \\[1em] \Rightarrow 3q = 13 \\[1em] \Rightarrow q = \dfrac{13}{3}.

p = 2 and q = 133\dfrac{13}{3}.

Hence, Option 3 is the correct option.

Question 23

If the line segment joining the points P and Q(3, -4) is bisected at the origin, then the coordinates of P are:

  1. (3, -2)

  2. (3, -4)

  3. (-3, -4)

  4. (-3, 4)

Answer

Let the coordinates of P be (x, y).

If the line segment joining the points P and Q(3, -4) is bisected at the origin, then the coordinates of P are: Reflection, RSA Mathematics Solutions ICSE Class 10.

Given, the origin (0, 0) is the mid-point of PQ.

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Substituting values we get :

⇒(0,0)=(x+32,y+(−4)2)⇒0=x+32 and 0=y−42⇒x+3=0 and y−4=0⇒x=−3 and y=4⇒P=(x,y)=(−3,4).\Rightarrow (0, 0) = \Big(\dfrac{x + 3}{2}, \dfrac{y + (-4)}{2}\Big) \\[1em] \Rightarrow 0 = \dfrac{x + 3}{2} \text{ and } 0 = \dfrac{y - 4}{2} \\[1em] \Rightarrow x + 3 = 0 \text{ and } y - 4 = 0 \\[1em] \Rightarrow x = -3 \text{ and } y = 4 \\[1em] \Rightarrow P = (x, y) = (-3, 4).

Hence, Option 4 is the correct option.

Question 24

A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, -5) is the mid-point of PQ, then the coordinates of P and Q are respectively:

  1. (0, 10) and (-4, 0)

  2. (0, -5) and (2, 0)

  3. (0, 4) and (-10, 0)

  4. (0, -10) and (4, 0)

Answer

Let P(0, a) be the point on y-axis and Q(b, 0) be the point on x-axis.

Given, (2, -5) is the mid-point of PQ.

A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, -5) is the mid-point of PQ, then the coordinates of P and Q are respectively: Reflection, RSA Mathematics Solutions ICSE Class 10.

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Substituting values we get :

⇒(2,−5)=(0+b2,a+02)⇒(2,−5)=(b2,a2)⇒2=b2 and −5=a2⇒b=4 and a=−10.\Rightarrow (2, -5) = \Big(\dfrac{0 + b}{2}, \dfrac{a + 0}{2}\Big) \\[1em] \Rightarrow (2, -5) = \Big(\dfrac{b}{2}, \dfrac{a}{2}\Big) \\[1em] \Rightarrow 2 = \dfrac{b}{2} \text{ and } -5 = \dfrac{a}{2} \\[1em] \Rightarrow b = 4 \text{ and } a = -10.

P(0, -10) and Q(4, 0)

Hence, Option 4 is the correct option.

Question 25

P is the mid-point of AB and Q is the mid-point of AP. If the coordinates of A are (2, -4) and coordinates of Q are (1, -1), then the coordinates of B are:

  1. (1, -1)

  2. (-1, 6)

  3. (-2, 6)

  4. (-2, 8)

Answer

Given,

Coordinates of A = (2, -4) and Q = (1, -1)

Let the coordinates of B be (a, b).

P is the mid-point of AB and Q is the mid-point of AP. If the coordinates of A are (2, -4) and coordinates of Q are (1, -1), then the coordinates of B are: Reflection, RSA Mathematics Solutions ICSE Class 10.

Given that Q is the mid-point of AP and P is the mid-point of AB.

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Let point P be (p,q)

Since Q is the mid-point of AP, we have

⇒(1,−1)=(2+p2,−4+q2)⇒1=2+p2 and −1=−4+q2⇒2+p=2 and −4+q=−2⇒p=2−2 and q=−2+4⇒p=0 and q=2.\Rightarrow (1, -1) = \Big(\dfrac{2 + p}{2}, \dfrac{-4 + q}{2}\Big) \\[1em] \Rightarrow 1 = \dfrac{2 + p}{2} \text{ and } -1 = \dfrac{-4 + q}{2} \\[1em] \Rightarrow 2 + p = 2 \text{ and } -4 + q = -2 \\[1em] \Rightarrow p = 2 - 2 \text{ and } q = -2 + 4 \\[1em] \Rightarrow p = 0 \text{ and } q = 2 .

Hence, the coordinates of P are (0, 2).

Now, since P is the mid-point of AB. We have,

⇒(0,2)=(2+a2,−4+b2)⇒0=2+a2 and 2=−4+b2⇒2+a=0 and −4+b=4⇒a=−2 and b=4+4⇒a=−2 and b=8.\Rightarrow (0, 2) = \Big(\dfrac{2 + a}{2}, \dfrac{-4 + b}{2}\Big) \\[1em] \Rightarrow 0 = \dfrac{2 + a}{2} \text{ and } 2 = \dfrac{-4 + b}{2} \\[1em] \Rightarrow 2 + a = 0 \text{ and } -4 + b = 4 \\[1em] \Rightarrow a = -2 \text{ and } b = 4 + 4 \\[1em] \Rightarrow a = -2 \text{ and } b = 8.

Therefore, the coordinates of B are (-2, 8).

Hence, Option 4 is the correct option.

Question 26

If a point R(235,335)R\Big(\dfrac{23}{5}, \dfrac{33}{5}\Big) divides the line segment PQ joining the points P(3, 5) and Q(x, y) in the ratio 2 : 3 internally, then the values of x and y respectively are :

  1. 4, 7

  2. 5, 9

  3. 7, 8

  4. 7, 9

Answer

Given,

Point R = (235,335)\Big(\dfrac{23}{5}, \dfrac{33}{5}\Big) and P(3, 5), Q(x, y).

divides the line segment PQ joining the points P(3, 5) and Q(x, y) in the ratio 2 : 3 internally, then the values of x and y respectively are :Reflection, RSA Mathematics Solutions ICSE Class 10.

Given,

m1 : m2 = 2 : 3

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒(235,335)=(2x+3×32+3,2y+3×52+3)⇒235=2x+95,335=2y+155⇒23=2x+9,33=2y+15⇒23−9=2x,33−15=2y⇒14=2x,18=2y⇒x=142,y=182⇒x=7,y=9.\Rightarrow \Big(\dfrac{23}{5}, \dfrac{33}{5}\Big) = \Big(\dfrac{2x + 3 \times 3}{2 + 3}, \dfrac{2y + 3 \times 5}{2 + 3}\Big) \\[1em] \Rightarrow \dfrac{23}{5} = \dfrac{2x + 9}{5}, \dfrac{33}{5} = \dfrac{2y + 15}{5} \\[1em] \Rightarrow 23 = 2x + 9, 33 = 2y + 15 \\[1em] \Rightarrow 23 - 9 = 2x, 33 - 15 = 2y \\[1em] \Rightarrow 14 = 2x, 18 = 2y \\[1em] \Rightarrow x = \dfrac{14}{2}, y = \dfrac{18}{2} \\[1em] \Rightarrow x = 7, y = 9.

Therefore, the values of x and y are 7 and 9 respectively.

Hence, Option 4 is the correct option.

Question 27

The points A, B and C divide the line segment joining the points P(-3, 8) and Q(9, -4) into four equal parts. If A is nearest to P, then the coordinates of A are:

  1. (-3, 5)

  2. (0, 5)

  3. (3, 5)

  4. (6, -1)

Answer

Given,

The points dividing PQ into four equal parts be A, B and C such that A is nearest to P.

Since the line segment PQ is divided into four equal parts, the point A divides PQ in the ratio 1 : 3.

The points A, B and C divide the line segment joining the points P(-3, 8) and Q(9, -4) into four equal parts. If A is nearest to P, then the coordinates of A are: Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values we get :

⇒(x,y)=(1×9+3×(−3)1+3,1×(−4)+3×81+3)⇒(9−94,−4+244)⇒(04,204)⇒(0,5).\Rightarrow (x, y) = \Big(\dfrac{1 \times 9 + 3 \times (-3)}{1 + 3}, \dfrac{1 \times (-4) + 3 \times 8}{1 + 3}\Big) \\[1em] \Rightarrow \Big(\dfrac{9 - 9}{4}, \dfrac{-4 + 24}{4}\Big) \\[1em] \Rightarrow \Big(\dfrac{0}{4}, \dfrac{20}{4}\Big) \\[1em] \Rightarrow (0, 5).

Hence, Option 2 is the correct option.

Question 28

The line 2x + y - 4 = 0 divides the line segment joining A(2, -2) and B(3, 7) in the ratio:

  1. 2 : 3

  2. 2 : 5

  3. 2 : 7

  4. 2 : 9

Answer

Let the required point be P(x, y) which divides A(2, -2) and B(3, 7) in the ratio k : 1.

The line 2x + y - 4 = 0 divides the line segment joining A(2, -2) and B(3, 7) in the ratio:Reflection, RSA Mathematics Solutions ICSE Class 10.

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values, we get :

⇒(x,y)=(k×3+1×2k+1,k×7+1×(−2)k+1)⇒(x,y)=(3k+2k+1,7k−2k+1)\Rightarrow (x, y) = \Big(\dfrac{k \times 3 + 1 \times 2}{k + 1}, \dfrac{k \times 7 + 1 \times (-2)}{k + 1}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{3k + 2}{k + 1}, \dfrac{7k - 2}{k + 1}\Big)

Since P lies on the line 2x + y - 4 = 0, substituting the values of x and y:

⇒2(3k+2k+1)+(7k−2k+1)−4=0⇒6k+4+7k−2k+1−4=0⇒13k+2k+1=4⇒13k+2=4(k+1)⇒13k+2=4k+4⇒9k=2⇒k=29⇒k:1=29:1=2:9.\Rightarrow 2\Big(\dfrac{3k + 2}{k + 1}\Big) + \Big(\dfrac{7k - 2}{k + 1}\Big) - 4 = 0 \\[1em] \Rightarrow \dfrac{6k + 4 + 7k - 2}{k + 1} - 4 = 0 \\[1em] \Rightarrow \dfrac{13k + 2}{k + 1} = 4 \\[1em] \Rightarrow 13k + 2 = 4(k + 1) \\[1em] \Rightarrow 13k + 2 = 4k + 4 \\[1em] \Rightarrow 9k = 2 \\[1em] \Rightarrow k = \dfrac{2}{9} \\[1em] \Rightarrow k : 1 = \dfrac{2}{9} : 1 = 2 : 9.

Hence, Option 4 is the correct option.

Question 29

The centre of the circle having end points of its one diameter as (-4, 2) and (4, -3) is:

  1. (0, -1)

  2. (2, -1)

  3. (0,−12)\Big(0, -\dfrac{1}{2}\Big)

  4. (4,−52)\Big(4, -\dfrac{5}{2}\Big)

Answer

Let the end points of the diameter be A(-4, 2) and B(4, -3). The centre of the circle is the mid-point of the diameter AB.

The centre of the circle having end points of its one diameter as (-4, 2) and (4, -3) is: Reflection, RSA Mathematics Solutions ICSE Class 10.

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Substituting values we get :

⇒(x,y)=(−4+42,2+(−3)2)=(02,−12)=(0,−12).\Rightarrow (x, y) = \Big(\dfrac{-4 + 4}{2}, \dfrac{2 + (-3)}{2}\Big) \\[1em] = \Big(\dfrac{0}{2}, \dfrac{-1}{2}\Big) \\[1em] = \Big(0, -\dfrac{1}{2}\Big).

Hence, Option 3 is the correct option.

Question 30

A circle has its centre at (4, 4). If one end of a diameter is (4, 0), then the coordinates of the other end are:

  1. (0, 4)

  2. (4, 8)

  3. (4, -8)

  4. (-4, -8)

Answer

Let one end of the diameter be A(4, 0) and the other end be B(x, y). Given that the centre of the circle is (4, 4), which is the mid-point of AB.

A circle has its centre at (4, 4). If one end of a diameter is (4, 0), then the coordinates of the other end are: Reflection, RSA Mathematics Solutions ICSE Class 10.

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Substituting values we get :

⇒(4,4)=(4+x2,0+y2)⇒4=4+x2 and 4=y2⇒x+4=8 and y=8⇒x=8−4=4 and y=8.\Rightarrow (4, 4) = \Big(\dfrac{4 + x}{2}, \dfrac{0 + y}{2}\Big) \\[1em] \Rightarrow 4 = \dfrac{4 + x}{2} \text{ and } 4 = \dfrac{y}{2} \\[1em] \Rightarrow x + 4 = 8 \text{ and } y = 8 \\[1em] \Rightarrow x = 8 - 4 = 4 \text{ and } y = 8.

The coordinates of the other end of the diameter are (4, 8).

Hence, Option 2 is the correct option.

Question 31

The vertices of a parallelogram in order are A(1, 2), B(4, y), C(x, 6) and D(3, 5). Then (x, y) is:

  1. (6, 3)

  2. (3, 6)

  3. (5, 6)

  4. (1, 4)

Answer

In a parallelogram, the diagonals bisect each other. Therefore, the mid-point of AC = mid-point of BD.

The vertices of a parallelogram in order are A(1, 2), B(4, y), C(x, 6) and D(3, 5). Then (x, y) is: Reflection, RSA Mathematics Solutions ICSE Class 10.

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Substituting values, we get :

For diagonal AC :

Mid-point of AC=(1+x2,2+62)=(1+x2,4)\text{Mid-point of AC} = \Big(\dfrac{1 + x}{2}, \dfrac{2 + 6}{2}\Big) = \Big(\dfrac{1 + x}{2}, 4\Big)

For diagonal BD:

Mid-point of BD=(4+32,y+52)=(72,y+52)\text{Mid-point of BD} = \Big(\dfrac{4 + 3}{2}, \dfrac{y + 5}{2}\Big) = \Big(\dfrac{7}{2}, \dfrac{y + 5}{2}\Big)

Since both mid-points are equal, we equate their coordinates:

⇒(1+x2,4)=(72,y+52)⇒1+x2=72,4=y+52⇒1+x=7,y+5=8⇒x=7−1,y=8−5⇒x=6,y=3.\Rightarrow \Big(\dfrac{1 + x}{2}, 4\Big) = \Big(\dfrac{7}{2}, \dfrac{y + 5}{2}\Big) \\[1em] \Rightarrow \dfrac{1 + x}{2} = \dfrac{7}{2}, 4 = \dfrac{y + 5}{2}\\[1em] \Rightarrow 1 + x = 7, y + 5 = 8 \\[1em] \Rightarrow x = 7 - 1, y = 8 - 5 \\[1em] \Rightarrow x = 6, y = 3.

(x, y) = (6, 3).

Hence, Option 1 is the correct option.

Question 32

The fourth vertex D of a parallelogram ABCD whose three vertices are A(-2, 3), B(6, 7) and C(8, 3) is:

  1. (0, 1)

  2. (0, -1)

  3. (-1, 0)

  4. (1, 0)

Answer

In a parallelogram, the diagonals bisect each other. Therefore, the mid-point of AC = mid-point of BD.

The fourth vertex D of a parallelogram ABCD whose three vertices are A(-2, 3), B(6, 7) and C(8, 3) is: Reflection, RSA Mathematics Solutions ICSE Class 10.

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Substituting values, we get :

For diagonal AC:

Mid-point of AC=(−2+82,3+32)=(62,62)=(3,3).\text{Mid-point of AC} = \Big(\dfrac{-2 + 8}{2}, \dfrac{3 + 3}{2}\Big) = \Big(\dfrac{6}{2}, \dfrac{6}{2}\Big) \\[1em] = (3, 3).

Let point D be (x, y).

For diagonal BD:

Mid-point of BD=(6+x2,7+y2)⇒(3,3)=(6+x2,7+y2)⇒3=6+x2 and 3=7+y2⇒6=6+x and 6=7+y⇒x=0 and y=6−7=−1.\text{Mid-point of BD} = \Big(\dfrac{6 + x}{2}, \dfrac{7 + y}{2}\Big) \\[1em] \Rightarrow (3, 3) = \Big(\dfrac{6 + x}{2}, \dfrac{7 + y}{2}\Big) \\[1em] \Rightarrow 3 = \dfrac{6 + x}{2} \text{ and } 3 = \dfrac{7 + y}{2} \\[1em] \Rightarrow 6 = 6 + x \text{ and } 6= 7 + y \\[1em] \Rightarrow x = 0 \text{ and } y = 6 - 7 = -1.

D = (x, y) = (0, -1).

Hence, Option 2 is the correct option.

Question 33

A(1, 4), B(4, 1) and C(x, 4) are the vertices of ΔABC. If the centroid of the triangle is G(4, 3), then x is equal to:

  1. 2

  2. 1

  3. 7

  4. 4

Answer

Given,

The vertices of the triangle be A(1, 4), B(4, 1) and C(x, 4). The centroid G of a triangle is given by the formula:

A(1, 4), B(4, 1) and C(x, 4) are the vertices of ΔABC. If the centroid of the triangle is G(4, 3), then x is equal to: Reflection, RSA Mathematics Solutions ICSE Class 10.

Centroid = (x1+x2+x33,y1+y2+y33)\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)

Substituting the given values we get :

⇒(4,3)=(1+4+x3,4+1+43)⇒(4,3)=(5+x3,93)⇒(4,3)=(5+x3,3)⇒4=5+x3⇒12=5+x⇒x=12−5=7.\Rightarrow (4, 3) = \Big(\dfrac{1 + 4 + x}{3}, \dfrac{4 + 1 + 4}{3}\Big) \\[1em] \Rightarrow (4, 3) = \Big(\dfrac{5 + x}{3}, \dfrac{9}{3}\Big) \\[1em] \Rightarrow (4, 3) = \Big(\dfrac{5 + x}{3}, 3\Big) \\[1em] \Rightarrow 4 = \dfrac{5 + x}{3} \\[1em] \Rightarrow 12 = 5 + x \\[1em] \Rightarrow x = 12 - 5 = 7.

Hence, Option 3 is the correct option.

Assertion Reason Questions

Question 1

Assertion (A): The coordinates of a point which divides a line segment joining the points (-3, 4) and (7, -6) in the ratio 1 : 2 internally are (13,23)\Big(\dfrac{1}{3}, \dfrac{2}{3}\Big).

Reason (R): The coordinates of the point which divides the line segment joining the points (x1, y1) and (x2, y2) internally in the ratio m : n are given by (mx2+nx1m+n,my2+ny1m+n)\Big(\dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n}\Big).

  1. A is true, R is false

  2. A is false, R is true

  3. Both A and R are true

  4. Both A and R are false

Answer

Let point P be (x, y), which divides line segment joining the points (-3, 4) and (7, -6) in the ratio 1 : 2.

The coordinates of a point which divides a line segment joining the points (-3, 4) and (7, -6) in the ratio 1 : 2. Reflection, RSA Mathematics Solutions ICSE Class 10.

Given,

m1 : m2 = 1 : 2

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substitute values we get:

⇒(x,y)=(1(7)+2(−3)1+2,1(−6)+2(4)1+2)=(7−63,−6+83)=(13,23).\Rightarrow (x, y) = \Big(\dfrac{1(7) + 2(-3)}{1 + 2}, \dfrac{1(-6) + 2(4)}{1 + 2}\Big) \\[1em] = \Big(\dfrac{7 - 6}{3}, \dfrac{-6 + 8}{3}\Big) \\[1em] = \Big(\dfrac{1}{3}, \dfrac{2}{3}\Big).

So, Assertion is true.

The coordinates of the point which divides the line segment joining the points (x1, y1) and (x2, y2) internally in the ratio m : n are given by (mx2+nx1m+n,my2+ny1m+n)\Big(\dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n}\Big).

So, reason is true.

Both A and R are true.

Hence, Option 3 is the correct option.

Question 2

Assertion (A): The coordinates of one end of a diameter of a circle are (1, 4). If its centre is at (2, -3), then the coordinates of the other end of the diameter are (-3, -10).

Reason (R): The centre of a circle is equidistant from each end of a diameter.

  1. A is true, R is false

  2. A is false, R is true

  3. Both A and R are true

  4. Both A and R are false

Answer

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

We know that,

Center is the mid-point of diameter of a circle.

Given,

Center = (2, -3)

One end of diameter = (1, 4)

Let other end of diameter be (a, b).

The coordinates of one end of a diameter of a circle are (1, 4). If its centre is at (2, -3), then the coordinates of the other end of the diameter are (-3, -10). Reflection, RSA Mathematics Solutions ICSE Class 10.

Substituting values, we get :

⇒(2,−3)=(1+a2,4+b2)⇒2=1+a2 and −3=4+b2⇒4=1+a and −6=4+b⇒4−1=a and −6−4=b⇒a=3 and b=−10.\Rightarrow (2, -3) = \Big(\dfrac{1 + a}{2}, \dfrac{4 + b}{2}\Big) \\[1em] \Rightarrow 2 = \dfrac{1 + a}{2} \text{ and } -3 = \dfrac{4 + b}{2} \\[1em] \Rightarrow 4 = 1 + a \text{ and } -6 = 4 + b \\[1em] \Rightarrow 4 - 1 = a \text{ and } -6 - 4 = b \\[1em] \Rightarrow a = 3 \text{ and } b = -10.

The correct coordinates of the other end are (3, -10).

So, Assertion (A) is false.

We know that,

Center is the mid-point of diameter of a circle.

Thus, centre of a circle is equidistant from each end of a diameter.

So, Reason (R) is true.

A is false, R is true.

Hence, Option 2 is the correct option.

Question 3

Assertion (A): If the points A(x, 1), B(8, 2), C(9, 4) and D(7, 3) are the vertices of a parallelogram, taken in order, then the value of x is 5.

Reason (R): Diagonals of a parallelogram bisect each other at right angles.

  1. A is true, R is false

  2. A is false, R is true

  3. Both A and R are true

  4. Both A and R are false

Answer

By mid-point formula,

(x, y) = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

If the points A(x, 1), B(8, 2), C(9, 4) and D(7, 3) are the vertices of a parallelogram, taken in order, then the value of x is 5. Reflection, RSA Mathematics Solutions ICSE Class 10.

Midpoint of AC:

MAC=(x+92,1+42)=(x+92,52)M_{AC} = \Big(\dfrac{x + 9}{2}, \dfrac{1 + 4}{2}\Big) = \Big(\dfrac{x + 9}{2}, \dfrac{5}{2}\Big)

Midpoint of BD:

MBD=(8+72,2+32)=(152,52)M_{BD} = \Big(\dfrac{8 + 7}{2}, \dfrac{2 + 3}{2}\Big) = \Big(\dfrac{15}{2}, \dfrac{5}{2}\Big)

Diagonals of //gm bisect each other.

Thus,

Mid-point of AC = Mid-point of BD

⇒(x+92,52)=(152,52)⇒x+92=152⇒x+9=15⇒x=15−9=6.\Rightarrow \Big(\dfrac{x + 9}{2}, \dfrac{5}{2}\Big) = \Big(\dfrac{15}{2}, \dfrac{5}{2}\Big) \\[1em] \Rightarrow \dfrac{x + 9}{2} = \dfrac{15}{2} \\[1em] \Rightarrow x + 9 = 15 \\[1em] \Rightarrow x = 15 - 9 = 6.

So, Assertion (A) is false.

Diagonals of a parallelogram bisect each other, but do NOT necessarily meet at right angles.

They are perpendicular only in special cases like a rhombus or a square.

So, Reason (R) is false.

Both A and R are false.

Hence, Option 4 is the correct option.

Question 4

Assertion (A): The coordinates of the centroid of a triangle whose vertices are (-1, -4), (4, 3) and (6, -2) are (3, -1).

Reason (R): If the vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the coordinates of its centroid are (x1+x2+x33,x1+x2+x33)\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{x_1 + x_2 + x_3}{3}\Big).

  1. A is true, R is false

  2. A is false, R is true

  3. Both A and R are true

  4. Both A and R are false

Answer

By using the centroid formula,

(x,y)=(x1+x2+x33,y1+y2+y33)(x, y) = \Big( \dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3} \Big)

The coordinates of the centroid of a triangle whose vertices are (-1, -4), (4, 3) and (6, -2) are (3, -1). Reflection, RSA Mathematics Solutions ICSE Class 10.

Substitute values we get,

⇒(x,y)=(−1+4+63,−4+3+(−2)3)⇒(x,y)=(93,−6+33)⇒(x,y)=(3,−33)⇒(x,y)=(3,−1).\Rightarrow (x,y) = \Big( \dfrac{-1 + 4 + 6}{3}, \dfrac{-4 + 3 + (-2)}{3} \Big) \\[1em] \Rightarrow (x,y) = \Big( \dfrac{9}{3}, \dfrac{-6 + 3}{3} \Big) \\[1em] \Rightarrow (x,y) = \Big( 3, \dfrac{-3}{3} \Big) \\[1em] \Rightarrow (x,y) = (3, -1).

So, Assertion (A) is true.

We know that,

If the vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the coordinates of its centroid are (x1+x2+x33,y1+y2+y33)\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big).

So, Reason (R) is false.

Hence, Option 1 is the correct option.

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