Chapter 11: Section Formula

Exercise 11

Question 1

Find the coordinates of the mid-points of the line segments joining the following pairs of points :

(i) (2, -3), (-6, 7)

(ii) (5, -11), (4, 3)

(iii) (a + 3, 5b), (2a - 1, 3b + 4).

Answer

(i) We know that,

Coordinates of mid-point of pair of points (x₁, y₁) and (x₂, y₂) is given by,

$\Rightarrow \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

∴ Mid-point of (2, -3), (-6, 7) is,

$\Rightarrow \Big(\dfrac{2 + (-6)}{2}, \dfrac{(-3) + (7)}{2}\Big) \\[1em] = \Big(-\dfrac{4}{2}, \dfrac{4}{2}\Big) \\[1em] = (-2, 2).$

Hence, (-2, 2) is the mid-point of (2, -3) and (-6, 7).

(ii) We know that,

Coordinates of mid-point of pair of points (x₁, y₁) and (x₂, y₂) is given by,

$\Rightarrow \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

∴ Mid-point of (5, -11), (4, 3) is,

$\Rightarrow \Big(\dfrac{5 + 4}{2}, \dfrac{(-11) + 3}{2}\Big) \\[1em] = \Big(\dfrac{9}{2}, -\dfrac{8}{2}\Big) \\[1em] = \Big(\dfrac{9}{2}, -4\Big).$

Hence, $(\dfrac{9}{2}, -4)$ is the mid-point of (5, -11) and (4, 3).

(iii) We know that,

Coordinates of mid-point of pair of points (x₁, y₁) and (x₂, y₂) is given by,

$\Rightarrow \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

∴ Mid-point of (a + 3, 5b), (2a - 1, 3b + 4) is,

$\Rightarrow \Big(\dfrac{a + 3 + 2a - 1}{2}, \dfrac{5b + 3b + 4}{2}\Big) \\[1em] = \Big(\dfrac{3a + 2}{2}, \dfrac{8b + 4}{2}\Big) \\[1em] = \Big(\dfrac{3a + 2}{2}, 4b + 2\Big).$

Hence, $(\dfrac{3a + 2}{2}, 4b + 2)$ is the mid-point of (a + 3, 5b) and (2a - 1, 3b + 4).

Question 2

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

By section-formula,

Point of division = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1+ m_ 2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Given,

m₁ : m₂ = 1 : 3

(x₁, y₁) = (-3, 9)

(x₂, y₂) = (1, -3)

Substituting values we get :

$\Rightarrow P = \Big(\dfrac{1 \times 1 + 3 \times -3}{1 + 3}, \dfrac{1 \times -3 + 3 \times 9}{1 + 3}\Big) \\[1em] \Rightarrow P = \Big(\dfrac{1 - 9}{4}, \dfrac{-3 + 27}{4}\Big) \\[1em] \Rightarrow P = \Big(\dfrac{-8}{4}, \dfrac{24}{4}\Big) \\[1em] \Rightarrow P = (-2, 6).$

Hence, the co-ordinates of P = (-2, 6).

Question 3(i)

Find the coordinates of the points of trisection of the line segment joining the points (3, -3) and (6, 9).

Answer

Let P(x₁, y₁) and Q(x₂, y₂) be the points of trisection of the points A(3, -3) and B(6, 9).

Find the coordinates of the points of trisection of the line segment joining the points (3, -3) and (6, 9). Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

AP = PQ = QB ⇒ 2AP = PB

⇒ $\dfrac{AP}{PB} = \dfrac{1}{2}$ ⇒ P divides AB in the ratio 1 : 2, so coordinates of P are

$\Rightarrow \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] = \Big(\dfrac{1 \times 6 + 2 \times 3}{1 + 2}, \dfrac{1 \times 9 + 2 \times (-3)}{1 + 2}\Big) \\[1em] = \Big(\dfrac{6 + 6}{3}, \dfrac{9 - 6}{3}\Big) \\[1em] = \Big(\dfrac{12}{3}, \dfrac{3}{3}\Big) \\[1em] = (4, 1).$

Q divides AB in the ratio 2 : 1, so coordinates of Q are

$\Rightarrow \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] = \Big(\dfrac{2 \times 6 + 1 \times 3}{2 + 1}, \dfrac{2 \times 9 + 1 \times (-3)}{2 + 1}\Big) \\[1em] = \Big(\dfrac{12 + 3}{3}, \dfrac{18 - 3}{3}\Big) \\[1em] = \Big(\dfrac{15}{3}, \dfrac{15}{3}\Big) \\[1em] = (5, 5).$

Hence, (4, 1) and (5, 5) are the coordinates of the points of trisection of the line segment joining the points (3, -3) and (6, 9).

Question 3(ii)

The line segment joining the points (3, -4) and (1, 2) is trisected at the points P and Q. If the coordinates of P and Q are (p, -2) and $\Big(\dfrac{5}{3}, q\Big)$ respectively, find the values of p and q.

Answer

Given P and Q trisect the points (3, -4) and (1, 2).

The line segment joining the points (3, -4) and (1, 2) is trisected at the points P and Q. If the coordinates of P and Q are (p, -2) and (5/3, q) respectively, find the values of p and q. Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

AP = PQ = QB ⇒ 2AP = PB

⇒ $\dfrac{AP}{PB} = \dfrac{1}{2}$ ⇒ P divides AB in the ratio 1 : 2, so coordinates of P are,

$\Rightarrow \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] = \Big(\dfrac{1 \times 1 + 2 \times 3}{1 + 2}, \dfrac{1 \times 2 + 2 \times (-4)}{1 + 2}\Big) \\[1em] = \Big(\dfrac{1 + 6}{3}, \dfrac{2 - 8}{3}\Big) \\[1em] = \Big(\dfrac{7}{3}, -\dfrac{6}{3}\Big) \\[1em] = \Big(\dfrac{7}{3}, -2\Big).$

Q divides AB in the ratio 2 : 1, so coordinates of Q are

$\Rightarrow \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] = \Big(\dfrac{2 \times 1 + 1 \times 3}{2 + 1}, \dfrac{2 \times 2 + 1 \times (-4)}{2 + 1}\Big) \\[1em] = \Big(\dfrac{2 + 3}{3}, \dfrac{4 - 4}{3}\Big) \\[1em] = \Big(\dfrac{5}{3}, 0\Big) \\[1em] = \Big(\dfrac{5}{3}, 0\Big).$

According to question,

Coordinates of P = (p, -2). Comparing it with $\Big(\dfrac{7}{3}, -2\Big)$ we get, p = $\dfrac{7}{3}.$

Coordinates of Q = $\Big(\dfrac{5}{3}, q\Big)$ . Comparing it with $\Big(\dfrac{5}{3}, 0\Big)$ we get, q = 0.

Hence, the value of p = $\dfrac{7}{3}$ and q = 0.

Question 4

The line segment joining the points A(3, 2) and B(5, 1) is divided at the points P in the ratio 1 : 2 and it lies on the line 3x - 18y + k = 0. Find the value of k.

Answer

Let (x, y) be the coordinates of the point P which divides the line segment joining A(3, 2) and B(5, 1) in the ratio 1 : 2.

∴ Coordinates of P are $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big).$

Putting the values from question in above formula we get,

$\Rightarrow \Big(\dfrac{1 \times 5 + 2 \times 3}{1 + 2}, \dfrac{1 \times 1 + 2 \times 2}{1 + 2}\Big) \\[1em] = \Big(\dfrac{5 + 6}{3}, \dfrac{1 + 4}{3}\Big) \\[1em] = \Big(\dfrac{11}{3}, \dfrac{5}{3}\Big).$

Since, P lies on the line 3x - 18y + k = 0.
∴ It will satisfy it.

$\Rightarrow 3\Big(\dfrac{11}{3}\Big) - 18\Big(\dfrac{5}{3}\Big) + k = 0 \\[1em] \Rightarrow 11 - 30 + k = 0 \\[1em] \Rightarrow -19 + k = 0 \\[1em] \Rightarrow k = 19.$

Hence, the value of k = 19.

Question 5

Find the coordinates of the point which is three-fourth of the way from A(3, 1) to B(-2, 5).

Answer

Let P(x, y) be the required point. Then according to question,

Find the coordinates of the point which is three-fourth of the way from A(3, 1) to B(-2, 5). Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

AP = $\dfrac{3}{4}$ AB

$\therefore \dfrac{AP}{AB} = \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{AP}{AP + PB} = \dfrac{3}{4} \\[1em] \Rightarrow 4AP = 3AP + 3PB \\[1em] \Rightarrow 4AP - 3AP = 3PB \\[1em] \Rightarrow AP = 3PB \\[1em] \Rightarrow \dfrac{AP}{PB} = \dfrac{3}{1} \\[1em] \therefore m_1 = 3, m_2 = 1.$

∴ By section formula coordinates of P are

$\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Putting the values from question in above formula we get,

$\Rightarrow \Big(\dfrac{3 \times (-2) + 1 \times 3}{3 + 1}, \dfrac{3 \times 5 + 1 \times 1}{3 + 1}\Big) \\[1em] = \Big(\dfrac{-6 + 3}{4}, \dfrac{15 + 1}{4}\Big) \\[1em] = \Big(-\dfrac{3}{4}, \dfrac{16}{4}\Big) \\[1em] = \Big(-\dfrac{3}{4}, 4\Big).$

Hence, the coordinates of P are $(-\dfrac{3}{4}, 4).$

Question 6

The line segment joining A(-3, 1) and B(5, -4) is a diameter of a circle whose centre is C. Find the coordinates of the point C.

Answer

Given, C is the centre of the circle and AB is the diameter.

C is the midpoint of AB.

Let coordinates of C be (x, y)

$\therefore x = \dfrac{x_1 + x_2}{2}, y = \dfrac{y_1 + y_2}{2} \\[1em] \Rightarrow x = \dfrac{-3 + 5}{2}, y = \dfrac{1 + (-4)}{2} \\[1em] \Rightarrow x = \dfrac{2}{2}, y = -\dfrac{3}{2} \\[1em] \Rightarrow x = 1, y = -\dfrac{3}{2}.$

Hence, coordinates of C are $(1, -\dfrac{3}{2})$ .

Question 7

The mid-point of the line segment joining the points (3m, 6) and (-4, 3n) is (1, 2m - 1). Find the values of m and n.

Answer

We know by mid-point formula,

$x = \dfrac{x_1 + x_2}{2} \text{ and } y = \dfrac{y_1 + y_2}{2}$

Given, the mid-point of (3m, 6) and (-4, 3n) is (1, 2m - 1).

$\therefore \text{x-coordinate } = 1 = \dfrac{3m + (-4)}{2} \\[1em] \Rightarrow 1 = \dfrac{3m - 4}{2} \\[1em] \Rightarrow 3m - 4 = 2 \\[1em] \Rightarrow 3m = 2 + 4 \\[1em] \Rightarrow 3m = 6 \\[1em] \Rightarrow m = 2.$

Similarly by midpoint formula,

$\text{y-coordinate } = 2m - 1 = \dfrac{6 + 3n}{2}$

Putting value of m = 2 in above equation we get,

$\Rightarrow 2 \times 2 - 1 = \dfrac{6 + 3n}{2} \\[1em] \Rightarrow 2(4 - 1) = 6 + 3n \\[1em] \Rightarrow 6 = 6 + 3n \\[1em] \Rightarrow 3n = 0 \\[1em] \Rightarrow n = 0.$

Hence, the value of m = 2 and n = 0.

Question 8

The coordinates of the mid-point of the line segment PQ are (1, -2). The coordinates of P are (-3, 2). Find the coordinates of Q.

Answer

Let the coordinates of Q be (x, y)

Coordinates of P are (-3, 2) and mid-point of PQ are (1, -2) then by mid-point formula, we get

$1 = \dfrac{(-3 + x)}{2} \text{ and } (-2) = \dfrac{(2 + y)}{2}$

⇒ -3 + x = 2 and 2 + y = -4
⇒ x = 2 + 3 and y = -4 - 2
⇒ x = 5 and y = -6.

Hence, coordinates of Q are (5, -6).

Question 9

AB is a diameter of a circle with centre C(-2, 5). If the point A is (3, -7). Find :

(i) The length of radius AC.

(ii) The coordinates of B.

Answer

Below figure shows the circle with Centre C(-2, 5)

AB is a diameter of a circle with centre C(-2, 5). If the point A is (3, -7). Find (i) The length of radius AC. (ii) The coordinates of B. Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

(i) We know that,

Distance formula = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

∴ The length of radius AC = $\sqrt{(-2 - 3)^2 + (5 - (-7))^2}$

$= \sqrt{(-5)^2 + (12)^2} \\[1em] = \sqrt{25 + 144} \\[1em] = \sqrt{169} \\[1em] = 13.$

Hence, the length of radius AC is 13 units.

(ii) Given, AB is the diameter and C is the mid-point

Let coordinates of B are (x, y) so, by mid-point formula,

$\dfrac{(3 + x)}{2} = -2 \text{ and } \dfrac{(y - 7)}{2} = 5$
⇒ 3 + x = -4 and y - 7 = 10
⇒ x = -4 - 3 and y = 10 + 7
⇒ x = -7 and y = 17.

Hence, coordinates of B are (-7, 17).

Question 10

Find the reflection (image) of the point (5, -3) in the point (-1, 3).

Answer

Let the coordinates of the image of the point A(5, -3) be A'(x, y) in the point (-1, 3) then,

The point (-1, 3) will be the mid-point of AA'

By midpoint formula,

$\Rightarrow -1 = \dfrac{(5 + x)}{2} \text{ and } 3 = \dfrac{(-3 + y)}{2}$
⇒ 5 + x = -2 and -3 + y = 6
⇒ x = -2 - 5 and y = 6 + 3
⇒ x = -7 and y = 9.

∴ A' = (x, y) = (-7, 9).

Hence, the coordinates of the images of (5, -3) in the point (-1, 3) is (-7, 9).

Question 11

The line segment joining A $\Big(-1, \dfrac{5}{3}\Big)$ and B(a, 5) is divided in the ratio 1 : 3 at P, the point where the line segment AB intersects y-axis. Calculate

(i) the value of a.

(ii) the coordinates of P.

Answer

(i) Let P(x, y) divide the line segment joining the points A $\Big(-1, \dfrac{5}{3}\Big)$ , B(a, 5) in the ratio 1 : 3

We know that,

Section-formula = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big).$

Putting values in above equation we get, coordinates of P

$= \Big(\dfrac{1 \times a + 3 \times (-1)}{1 + 3}, \dfrac{1 \times 5 + 3 \times \dfrac{5}{3}}{1 + 3}\Big) \\[1em] = \Big(\dfrac{a - 3}{4}, \dfrac{5 + 5}{4}\Big) \\[1em] = \Big(\dfrac{a - 3}{4}, \dfrac{10}{4}\Big) \\[1em] = \Big(\dfrac{a - 3}{4}, \dfrac{5}{2}\Big).$

According to the question,

The point P lies on y-axis so, x coordinate will be equal to zero.

$\therefore \dfrac{a - 3}{4} = 0$
⇒ a - 3 = 0
⇒ a = 3.

Hence, the value of a = 3.

(ii) From part (i) we get the y coordinate of P = $\dfrac{5}{2}$ .

Hence, the coordinates of P are $\Big(0, \dfrac{5}{2}\Big).$

Question 12

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

Answer

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

As the point P(-4, 1) divides the line segment joining the points A (2, -2) and B (x, y) in the ratio 3 : 5, we have

We know that,

Section-formula = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big).$

Putting values in above equation we get coordinates of P as

$= \Big(\dfrac{3 \times x + 5 \times 2}{3 + 5}, \dfrac{3 \times y + 5 \times (-2)}{3 + 5}\Big) \\[1em] = \Big(\dfrac{3x + 10}{8}, \dfrac{3y - 10}{8}\Big).$

According to question the coordinates of P are (-4, 1) comparing we get,

$\Rightarrow \dfrac{3x + 10}{8} = -4 \text{ and } \dfrac{3y - 10}{8} = 1$
⇒ 3x + 10 = -32 and 3y - 10 = 8
⇒ 3x = -32 - 10 and 3y = 8 + 10
⇒ 3x = -42 and 3y = 18
⇒ x = -14 and y = 6.

∴ B = (x, y) = (-14, 6).

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

Question 13(i)

In what ratio does the point (5, 4) divide the line segment joining the points (2, 1) and (7, 6) ?

Answer

Let the point P(5, 4) divide the line segment joining the points (2, 1), (7, 6) in the ratio m₁ : m₂.

By Section-formula, we get the coordinates of point P as:

$\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big).$

Putting values in x coordinate of above equation we get,

$= \dfrac{m_1 \times 7 + m_2 \times 2}{m_1 + m_2} \\[1em] = \dfrac{7m_1 + 2m_2}{m_1 + m_2} \\[1em]$

According to question, the x-coordinate of P = 5. Comparing we get,

$\Rightarrow \dfrac{7m_1 + 2m_2}{m_1 + m_2} = 5 \\[1em] \Rightarrow 7m_1 + 2m_2 = 5m_1 + 5m_2 \\[1em] \Rightarrow 7m_1 - 5m_1 = 5m_2 - 2m_2 \\[1em] \Rightarrow 2m_1 = 3m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{3}{2}.$

Hence, the ratio in which point (5, 4) divides the line segment is 3 : 2.

Question 13(ii)

In what ratio does the point (-4, b) divide the line segment joining the points P(2, -2), Q(-14, 6)? Hence, find the value of b.

Answer

Let the point P(-4, b) divide the line segment joining the points P(2, -2) and Q(-14, 6) in the ratio m₁ : m₂.

By section formula, the x-coordinate = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}\Big)$

Putting value in above formula we get,

$\Rightarrow -4 = \dfrac{m_1 \times -14 + m_2 \times 2}{m_1 + m_2} \\[1em] \Rightarrow -4m_1 - 4m_2 = -14m_1 + 2m_2 \\[1em] \Rightarrow -4m_1 + 14m_1 = 2m_2 + 4m_2 \\[1em] \Rightarrow 10m_1 = 6m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{6}{10} \\[1em] \Rightarrow m_1 : m_2 = 3 : 5.$

By section formula, y-coordinate = $\Big(\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Putting value in above formula we get,

$\Rightarrow b = \dfrac{m_1 \times 6 + m_2 \times -2}{m_1 + m_2} \\[1em] = \dfrac{6m_1 - 2m_2}{m_1 + m_2} \\[1em] = \dfrac{6 \times 3 - 2 \times 5}{3 + 5} \\[1em] = \dfrac{18 - 10}{8} \\[1em] = \dfrac{8}{8} \\[1em] = 1.$

Hence, the ratio is 3 : 5 and the value of b = 1.

Question 14

The line segment joining A(2, 3) and B(6, -5) is intersected by x-axis at a point K. Write down the ordinate of the point K. Hence, find the ratio in which K divides AB. Also find the coordinates of point K.

Answer

Let the coordinates of K be (x, 0) as it intersects x-axis. Let point K divides the line segment joining the points
A(2, 3) and B(6, -5) in the ratio m₁ : m₂.

By section formula, y-coordinate = $\Big(\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Putting value in above formula we get,

$\Rightarrow 0 = \dfrac{m_1 \times (-5) + m_2 \times 3}{m_1 + m_2} \\[1em] \Rightarrow 0 = -5m_1 + 3m_2 \\[1em] \Rightarrow 5m_1 = 3m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{3}{5}$

Now for x coordinate by section formula we get,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 6 + 5 \times 2}{3 + 5} \\[1em] = \dfrac{18 + 10}{8} \\[1em] = \dfrac{28}{8} \\[1em] = \dfrac{7}{2}.$

∴ K = (x, 0) = $\Big(\dfrac{7}{2}, 0\Big).$

Hence, the coordinates of K are $\Big(\dfrac{7}{2}, 0\Big)$ and the ratio is 3 : 5.

Question 15

If A = (-4, 3) and B = (8, -6),

(i) find the length of AB.

(ii) in what ratio is the line segment joining AB, divided by the x-axis?

Answer

The graph is given below:

If A = (-4, 3) and B = (8, -6), (i) find the length of AB. (ii) in what ratio is the line segment joining AB, divided by the x-axis? Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

(i) Given,

A = (-4, 3), B = (8, -6)

$\text{Length of AB = } \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] = \sqrt{(8 - (-4))^2 + (-6 - 3)^2} \\[1em] = \sqrt{(8 + 4)^2 + (-6 - 3)^2} \\[1em] = \sqrt{12^2 + (-9)^2} \\[1em] = \sqrt{144 + 81} \\[1em] = \sqrt{225} \\[1em] = 15.$

Hence, the length of AB = 15 units.

(ii) From graph, we see that O (0, 0) lies on AB.
Let O divide AB in the ratio m₁ : m₂.

By section-formula the coordinates of point dividing a line in m₁ : m₂ are given by

$\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Putting values in above equation for finding x coordinates we get,

$\Rightarrow \dfrac{m_1 \times 8 + m_2 \times -4}{m_1 + m_2} \\ \\[1em] \Rightarrow \dfrac{8m_1 - 4m_2}{m_1 + m_2}.$

Since origin (0, 0) is the dividing point the x-coordinate = 0. Comparing it with above equation we get,

$\dfrac{8m_1 - 4m_2}{m_1 + m_2} = 0 \\[1em] 8m_1 - 4m_2 = 0 \\[1em] 8m_1 = 4m_2 \\[1em] \dfrac{m_1}{m_2} = \dfrac{4}{8} \\[1em] \dfrac{m_1}{m_2} = \dfrac{1}{2}.$

∴ m₁ : m₂ = 1 : 2.

Hence, the x-axis divides the line segment AB in the ratio 1 : 2.

Question 16

In what ratio does the line x - y - 2 = 0 divide the line segment joining the points (3, -1) and (8, 9)? Also, find the coordinates of the point of division.

Answer

Let the point A be (3, -1) and point B be (8, 9) and let the line x - y - 2 = 0 divide the line segment AB in the ratio m₁ : m₂ at point P(x, y) then,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{m_1 \times 8 + m_2 \times 3}{m_1 + m_2} \\[1em] = \dfrac{8m_1 + 3m_2}{m_1 + m_2} \qquad \text{....[Eq 1]} \\[1em] \text{and y } = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{m_1 \times 9 + m_2 \times -1}{m_1 + m_2} \\[1em] = \dfrac{9m_1 - m_2}{m_1 + m_2} \qquad \text{....[Eq 2]}.$

Given, the point P(x, y) lies on the line x - y - 2 = 0.

$\therefore \dfrac{8m_1 + 3m_2}{m_1 + m_2} - \dfrac{9m_1 - m_2}{m_1 + m_2} - 2 = 0 \\[1em] \Rightarrow \dfrac{8m_1 + 3m_2 - 9m_1 + m_2 - 2m_1 - 2m_2}{m_1 + m_2} = 0 \\[1em] \Rightarrow -3m_1 + 2m_2 = 0 \\[1em] \Rightarrow 3m_1 = 2m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{2}{3} \\[1em].$

Putting value of m₁ : m₂ in Eq 1,

$\Rightarrow x = \dfrac{8 \times 2 + 3 \times 3}{2 + 3} \\[1em] = \dfrac{16 + 9}{5} \\[1em] = \dfrac{25}{5} \\[1em] = 5.$

Putting value of m₁ : m₂ in Eq 2,

$\Rightarrow y = \dfrac{9 \times 2 - 3}{2 + 3} \\[1em] = \dfrac{18 - 3}{5} \\[1em] = \dfrac{15}{5} \\[1em] = 3.$

Hence, the ratio is 2 : 3 and coordinates of P are (5, 3).

Question 17

Given, a line segment AB joining the points A(-4, 6) and B(8, -3). Find :

(i) The ratio in which AB is divided by the y-axis.

(ii) The coordinates of the point of intersection.

(iii) The length of AB.

Answer

(i) Let the y-axis divide AB in the ratio m₁ : m₂.

By section formula, the x-coordinate = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}\Big)$

Since, the x coordinate on y-axis is 0. Putting value in above formula we get,

$\Rightarrow 0 = \dfrac{m_1 \times 8 + m_2 \times (-4)}{m_1 + m_2} \\[1em] \Rightarrow 8m_1 - 4m_2 = 0 \\[1em] \Rightarrow 8m_1 = 4m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{4}{8} \\[1em] \Rightarrow m_1 : m_2 = 1 : 2$

Hence, required ratio = $\dfrac{1}{2} : 1$ or 1 : 2.

(ii) By section formula, the y-coordinate is given by

$\Big(\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Putting value in above formula we get,

$\Rightarrow y = \dfrac{1 \times -3 + 2 \times 6}{1 + 2} \\[1em] = \dfrac{-3 + 12}{3} \\[1em] = \dfrac{9}{3} \\[1em] = 3.$

Hence, the coordinates of the point of intersection are (0, 3).

(iii) Distance formula = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

By distance formula,

$AB = \sqrt{(8 - (-4))^2 + (-3 - 6)^2} \\[1em] = \sqrt{12^2 + (-9)^2} \\[1em] = \sqrt{144 + 81} \\[1em] = \sqrt{225} \\[1em] = 15.$

Hence, the length of AB = 15 units.

Question 18

Calculate the length of the median through the vertex A of the triangle ABC with vertices A(7, -3), B(5, 3) and C(3, -1).

Answer

Below figure shows the triangle ABC with vertices A(7, -3), B(5, 3) and C(3, -1):

Calculate the length of the median through the vertex A of the triangle ABC with vertices A(7, -3), B(5, 3) and C(3, -1). Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Let D (x, y) be the mid-point of BC, then AD is the median through A.

As D is the mid-point of BC, then coordinates of D by mid-point formula are

$\Rightarrow x = \dfrac{5 + 3}{2} \text{ and } y = \dfrac{3 + (-1)}{2} \\[1em] \Rightarrow x = \dfrac{8}{2} \text{ and } y = \dfrac{2}{2} \\[1em] \Rightarrow x = 4 \text{ and } y = 1.$

Hence, coordinates of D are (4, 1).

Distance formula = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

By distance formula length of median,

$AD = \sqrt{(7 - 4)^2 + (-3 - 1)^2} \\[1em] = \sqrt{3^2 + (-4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5.$

Hence, the length of the median through vertex A is 5 units.

Question 19

Three consecutive vertices of a parallelogram ABCD are A(1, 2), B(1, 0) and C(4, 0). Find the fourth vertex D.

Answer

In the parallelogram ABCD, the three consecutive vertices are A(1, 2), B(1, 0) and C(4, 0). Let the fourth vertex D be (x, y) as shown in the figure below:

Three consecutive vertices of a parallelogram ABCD are A(1, 2), B(1, 0) and C(4, 0). Find the fourth vertex D. Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Let O be the mid-point of AC, the diagonal of ABCD.

By Mid-point formula, the coordinates of O = $\Big(\dfrac{1 + 4}{2}, \dfrac{2 + 0}{2}\Big)$

∴ Coordinates of O will be $\Big(\dfrac{5}{2}, 1\Big)$

The mid-point of BD is $\Big(\dfrac{1 + x}{2}, \dfrac{y}{2}\Big)$

Since, the diagonals of a parallelogram bisect each other, the mid-points of AC and BD are same.

$\therefore \dfrac{5}{2} = \dfrac{1 + x}{2} \text{ and } 1 = \dfrac{0 + y}{2} \\[1em] \Rightarrow 5 = 1 + x \text{ and } y + 0 = 2 \\[1em] \Rightarrow x = 5 - 1 \text{ and } y = 2. \\[1em] \Rightarrow x = 4 \text{ and } y = 2.$

Hence, coordinates of D are (4, 2).

Question 20

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

Answer

A(-2, -1), B(1, 0), C(p, 3) and D(1, q) are the vertices of a parallelogram ABCD as shown in the figure below:

If the points A(-2, -1), B(1, 0), C(p, 3) and D(1, q) form a parallelogram ABCD, find the values of p and q. Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

∴ Diagonal AC and BD bisect each other at O.
O is the mid-point of AC as well as BD. Let the coordinates of O be (x, y).

When O is the mid-point of AC, then by mid-point formula

$\Rightarrow x = \dfrac{p + (-2)}{2} \text{ and y = } \dfrac{3 + (-1)}{2} \\[1em] \Rightarrow x = \dfrac{p - 2}{2} \text{ and y = } \dfrac{2}{2} \\[1em] \therefore x = \dfrac{p - 2}{2} \text{ and } y = 1.$

When O is the mid-point of BD, then by mid-point formula

$\Rightarrow x = \dfrac{1 + 1}{2} \text{ and y = } \dfrac{0 + q}{2} \\[1em] \Rightarrow x = \dfrac{2}{2} \text{ and y = } \dfrac{q}{2} \\[1em] \therefore x = 1 \text{ and } y = \dfrac{q}{2}.$

Now comparing, we get the values of the coordinates of point O in both cases,

$\Rightarrow \dfrac{p - 2}{2} = 1 \text{ and } \dfrac{q}{2} = 1 \\[1em] \Rightarrow p - 2 = 2 \text{ and } q = 2 \\[1em] \Rightarrow p = 4 \text{ and } q = 2.$

Hence, the value of p = 4 and q = 2.

Question 21

If two vertices of a parallelogram are (3, 2), (-1, 0) and its diagonals meet at (2, -5), find the other two vertices of the parallelogram.

Answer

The mid-point of the line segment joining the points (3, 2) and (-1, 0) is $\Big(\dfrac{3 + (-1)}{2}, \dfrac{2 + 0}{2}\Big)$ = (1, 1) which is not the same as (2, -5), therefore, the given points cannot be the opposite vertices. Hence, these vertices are adjoining.

The below figure shows the parallelogram:

If two vertices of a parallelogram are (3, 2), (-1, 0) and its diagonals meet at (2, -5), find the other two vertices of the parallelogram. Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Let coordinates of C be (x, y) then by mid-point formula,

$\Rightarrow 2 = \dfrac{x + 3}{2} \text{ and} -5 = \dfrac{y + 2}{2}$
⇒ x + 3 = 4 and y + 2 = -10
⇒ x = 1 and y = -12.

∴ Coordinates of C are (1, -12).

Now finding coordinates of D, let D be (m, n). Applying mid-point formula we get,

$\Rightarrow 2 = \dfrac{m - 1}{2} \text{ and } -5 = \dfrac{n + 0}{2}$
⇒ m - 1 = 4 and n = -10
⇒ m = 5 and n = -10.

∴ Coordinates of D are (5, -10).

Hence, the coordinates of C are (1, -12) and D are (5, -10).

Question 22

Find the third vertex of a triangle if its two vertices are (-1, 4) and (5, 2) and mid-point of one side is (0, 3).

Answer

Let A(-1, 4) and B(5, 2) be the two points and let C(x, y) be the third vertex of the triangle as shown in the figure below:

Find the third vertex of a triangle if its two vertices are (-1, 4) and (5, 2) and mid-point of one side is (0, 3). Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

By mid-point formula, mid-point of AB

$= \Big(\dfrac{-1 + 5}{2}, \dfrac{4 + 2}{2}\Big) \\[1em] = \Big(\dfrac{4}{2}, \dfrac{6}{2}\Big) \\[1em] = (2, 3)$

Since, (0, 3) is not the mid-point of AB hence, it is either the mid point of AC or BC.

Let D(0, 3) be the midpoint of AC. So, by mid-point formula,

$0 = \dfrac{x + (-1)}{2} \text{ and } 3 = \dfrac{y + 4}{2}$
⇒ x - 1 = 0 and y + 4 = 6
⇒ x = 1 and y = 2.

∴ Coordinates of C will be (1, 2).

Let D(0, 3) be the midpoint of BC. So, by mid-point formula,

$0 = \dfrac{5 + x}{2} \text{ and } 3 = \dfrac{2 + y}{2}$
⇒ x + 5 = 0 and y + 2 = 6
⇒ x = -5 and y = 4.

∴ Coordinates of C will be (-5, 4).

Hence, the coordinates of C will be (1, 2) or (-5, 4).

Question 23

Find the coordinates of the vertices of the triangle, the middle points of whose sides are $\Big(0, \dfrac{1}{2}\Big), \Big(\dfrac{1}{2}, \dfrac{1}{2}\Big) \text{ and } \Big(\dfrac{1}{2}, 0\Big)$ .

Answer

Let ABC be a triangle in which D, E and F are the mid-points of sides AB, BC and CA respectively.

Let coordinates of A be (x₁, y₁), B(x₂, y₂), C(x₃, y₃).

Find the coordinates of the vertices of the triangle, the middle points of whose sides are (0, 1/2), (1/2, 1/2) and (1/2,0). Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Applying mid-point formula on side AB,

$0 = \dfrac{x_1 + x_2}{2} \\[1em] \Rightarrow x_1 + x_2 = 0 \qquad \text{....[Eq 1]} \\[1.5em] \dfrac{1}{2} = \dfrac{y_1 + y_2}{2} \\[1em] \Rightarrow y_1 + y_2 = 1 \qquad \text{....[Eq 2]} \\[1em]$

Applying mid-point formula on side BC,

$\dfrac{1}{2} = \dfrac{x_2 + x_3}{2} \\[1em] \Rightarrow x_2 + x_3 = 1. \qquad \text{....[Eq 3]} \\[1em] \dfrac{1}{2} = \dfrac{y_2 + y_3}{2} \\[1em] \Rightarrow y_2 + y_3 = 1 \qquad \text{....[Eq 4]} \\[1em]$

Applying mid-point formula on side CA,

$\dfrac{1}{2} = \dfrac{x_3 + x_1}{2} \\[1em] \Rightarrow x_3 + x_1 = 1. \qquad \text{....[Eq 5]} \\[1em] 0 = \dfrac{y_3 + y_1}{2} \\[1em] \Rightarrow y_3 + y_1 = 0 \qquad \text{....[Eq 6]} \\[1em]$

Adding Eq 1, 3 and 5,

$x_1 + x_2 + x_2 + x_3 + x_3 + x_1 = 0 + 1 + 1.$
∴ x₁ + x₂ + x₃ = 1. $\qquad \text{....[Eq 7]}$

On subtracting Eq 3 from Eq 7 we get,
x₁ + x₂ + x₃ - (x₂ + x₃) = 1 - 1
x₁ + x₂ + x₃ - x₂ - x₃ = 0
x₁ = 0.

On subtracting Eq 5 from Eq 7 we get,
x₁ + x₂ + x₃ - (x₃ + x₁) = 1 - 1
x₁ + x₂ + x₃ - x₃ - x₁ = 0
x₂ = 0.

On subtracting Eq 1 from Eq 7 we get,
x₁ + x₂ + x₃ - (x₁ + x₂) = 1 - 0
x₁ + x₂ + x₃ - x₁ - x₂ = 1
x₃ = 1.

Adding Eq 2, 4 and 5,

$y_1 + y_2 + y_2 + y_3 + y_3 + y_1 = 1 + 1 + 0.$
∴ y₁ + y₂ + y₃ = 1. (Eq 8)

On subtracting Eq 4 from Eq 8 we get,
y₁ + y₂ + y₃ - (y₂ + y₃) = 1 - 1
y₁ + y₂ + y₃ - y₂ - y₃ = 0
y₁ = 0.

On subtracting Eq 6 from Eq 8 we get,
y₁ + y₂ + y₃ - (y₃ + y₁) = 1 - 0
y₁ + y₂ + y₃ - y₃ - y₁ = 1
y₂ = 1.

On subtracting Eq 2 from Eq 8 we get,
y₁ + y₂ + y₃ - (y₁ + y₂) = 1 - 1
y₁ + y₂ + y₃ - y₁ - y₂ = 1 - 1
y₃ = 0.

Hence, coordinates of vertices of triangle are (0, 0), (0, 1) and (1, 0) respectively.

Question 24

Show by section formula that the points (3, -2), (5, 2) and (8, 8) are collinear.

Answer

Let the point (5, 2) divides the line joining the points (3, -2) and (8, 8) in the ratio m₁ : m₂.

By section formula the x-coordinate of dividing points is given by

$\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}\Big)$

Putting values we get,

$5 = \dfrac{m_1 \times 8 + m_2 \times 3}{m_1 + m_2} \\[1em] \Rightarrow 8m_1 + 3m_2 = 5m_1 + 5m_2 \\[1em] \Rightarrow 8m_1 - 5m_1 = 5m_2 - 3m_2 \\[1em] \Rightarrow 3m_1 = 2m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{2}{3}. \qquad \text{....[Eq 1]}$

By section formula the y-coordinate of dividing points is given by

$\Big(\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Putting values we get,

$2 = \dfrac{m_1 \times 8 + m_2 \times (-2)}{m_1 + m_2} \\[1em] \Rightarrow 8m_1 - 2m_2 = 2m_1 + 2m_2 \\[1em] \Rightarrow 8m_1 - 2m_1 = 2m_2 + 2m_2 \\[1em] \Rightarrow 6m_1 = 4m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{4}{6} \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{2}{3} \qquad \text{....[Eq 2]}$

Since, we get ratio m₁ : m₂ from both the equations it means that the point (5, 2) lies on the line joining the points (3, -2) and (8, 8).

Hence, proved that the points (3, -2), (5, 2) and (8, 8) are collinear.

Question 25

Find the value of p for which the points (-5, 1), (1, p) and (4, -2) are collinear.

Answer

Since, the points A, B are C are collinear then the point A(-5, 1) divides BC in the ratio of m₁ : m₂.

For x-coordinate using section formula,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow -5 = \dfrac{m_1 \times 4 + m_2 \times 1}{m_1 + m_2} \\[1em] \Rightarrow -5 = \dfrac{4m_1 + m_2}{m_1 + m_2} \\[1em] \Rightarrow -5m_1 - 5m_2 = 4m_1 + m_2 \\[1em] \Rightarrow -5m_1 - 4m_1 = m_2 + 5m_2 \\[1em] \Rightarrow -9m_1 = 6m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = -\dfrac{6}{9} = -\dfrac{2}{3} \qquad \text{....[Eq 1]}$

For y-coordinate using section formula,

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow 1 = \dfrac{m_1 \times (-2) + m_2 \times p}{m_1 + m_2} \\[1em] \Rightarrow 1 = \dfrac{-2m_1 + m_2p}{m_1 + m_2} \\[1em] \Rightarrow m_1 + m_2 = -2m_1 + m_2p \\[1em] \Rightarrow m_1 + 2m_1 = m_2p - m_2 \\[1em] \Rightarrow 3m_1 = m_2(p - 1) \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{p - 1}{3} \qquad \text{....[Eq 2]}$

Comparing Eq 1 and 2,

$\Rightarrow -\dfrac{2}{3} = \dfrac{p - 1}{3} \\[1em] \Rightarrow -\dfrac{2}{3} = \dfrac{p - 1}{3} \\[1em] \Rightarrow -2 \times 3 = 3(p - 1) \\[1em] \Rightarrow -6 = 3p - 3 \\[1em] \Rightarrow 3p = -3 \\[1em] \Rightarrow p = -1.$

Hence, the value of p = -1.

Question 26

The mid-point of the line segment AB shown in the adjoining diagram is (4, -3). Write down the coordinates of A and B.

Answer

From graph we see,

A lies on x-axis and B be on the y-axis.
Let coordinates of A be (x, 0) and of B be (0, y). P(4, -3) is the mid-point of AB.

$\therefore 4 = \dfrac{x + 0}{2} \text{ and } -3 = \dfrac{0 + y}{2}$
⇒ x = 8 and y = -6.

Hence, coordinates of A will be (8, 0) and of B will be (0, -6).

Question 27

Find the coordinates of the centroid of a triangle whose vertices are :

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

Answer

Coordinates of the centroid of the triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃) is given by

$\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)$

Putting values in the formula we get,

$= \Big(\dfrac{-1 + 1 + 5}{3}, \dfrac{3 + (-1) + 1}{3}\Big) \\[1em] = \Big(\dfrac{5}{3}, \dfrac{3}{3}\Big) \\[1em] = \Big(\dfrac{5}{3}, 1\Big).$

Hence, the coordinates of the centroid of the triangle are $\Big(\dfrac{5}{3}, 1\Big)$ .

Question 28

Two vertices of a triangle are (3, -5) and (-7, 4). Find the third vertex given that the centroid is (2, -1).

Answer

Let the coordinates of third vertex be (x, y) and other two vertices are (3, -5) and (-7, 4) and centroid = (2, -1).

Coordinates of Centroid of the triangle are given by

$\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big) \\[1em] \Rightarrow 2 = \dfrac{3 + (-7) + x}{3} \text{ and } -1 = \dfrac{-5 + 4 + y}{3} \\[1em] \Rightarrow 6 = x - 4 \text{ and } -3 = y - 1 \\[1em] \Rightarrow x = 6 + 4 \text{ and } y = -3 + 1 \\[1em] \Rightarrow x = 10 \text{ and } y = -2.$

Hence, the coordinates of centroid are (10, -2).

Question 29

The vertices of a triangle are A(-5, 3), B(p, -1) and C(6, q). Find the values of p and q if the centroid of the triangle ABC is the point (1, -1).

Answer

The vertices of the triangle ABC are A(-5, 3), B(p, -1), C(6, q) and the centroid of the △ABC will be

$\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)$

Putting the values in above formula,

$= \Big(\dfrac{-5 + p + 6}{3}, \dfrac{3 + (-1) + q}{3}\Big) \\[1em] = \Big(\dfrac{1 + p}{3}, \dfrac{2 + q}{3}\Big).$

Given, the centroid of the triangle is (1, -1).

Comparing we get,

$\dfrac{1 + p}{3} = 1 \text{ and } \dfrac{q + 2}{3} = -1$
⇒ 1 + p = 3 and 2 + q = -3
⇒ p = 3 - 1 and q = -3 - 2
⇒ p = 2 and q = -5.

Hence, the values are p = 2 and q = -5.

Multiple Choice Questions

Question 1

The points A(9, 0), B(9, 6), C(-9, 6) and D(-9, 0) are the vertices of a

rectangle
square
rhombus
trapezium

Answer

We know that,

Distance-Formula = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given, A(9, 0), B(9, 6), C(-9, 6) and D(-9, 0)

Calculating the length of sides by distance formula.

$AB = \sqrt{(9 - 9)^2 + (6 - 0)^2} \\[1em] = \sqrt{0 + 36} \\[1em] = 6 \text{ units.} \\[1em] BC = \sqrt{(-9 - 9)^2 + (6 - 6)^2} \\[1em] = \sqrt{(-18)^2 + 0^2} \\[1em] = \sqrt{324} \\[1em] = 18 \text{ units.} \\[1em] CD = \sqrt{(-9 - (-9))^2 + (0 - 6)^2} \\[1em] = \sqrt{0 + 36} \\[1em] = \sqrt{36} \\[1em] = 6 \text{ units.} \\[1em] DA = \sqrt{(9 - (-9))^2 + (0 - 0)^2} \\[1em] = \sqrt{18^2 + 0} \\[1em] = \sqrt{324} \\[1em] = 18 \text{ units.}$

Since, AB = CD and BC = DA means opposite sides are equal.

Hence, Option 1 is the correct option.

Question 2

If $P(\dfrac{a}{3}, 4)$ is the mid-point of the line segment joining the points Q(-6, 5) and R(-2, 3), then the value of a is

-4
-6
12
-12

Answer

We know that,

Mid-point formula = $\Big(\dfrac{(x_1 + x_2)}{2}, \dfrac{(y_1 + y_2)}{2}\Big)$

Given, P is the midpoint of line segment QR. Putting the values in mid-point formula for x-coordinate we get,

$\Rightarrow \dfrac{a}{3} = \dfrac{-6 + (-2)}{2} \\[1em] \Rightarrow a = \dfrac{-8}{2} \times 3 \\[1em] \Rightarrow a = -4 \times 3 \\[1em] \Rightarrow a = -12.$

Hence, Option 4 is the correct option.

Question 3

If the end points of a diameter of a circle are A(-2, 3) and B(4, -5) then the coordinates of its centre are

(2, -2)
(1, -1)
(-1, 1)
(-2, 2)

Answer

We know that,

Mid-point formula = $\Big(\dfrac{(x_1 + x_2)}{2}, \dfrac{(y_1 + y_2)}{2}\Big)$

Given A and B are end points of the diameter so by mid-point formula the coordinates of the centre of the circle are,

$\Rightarrow \Big(\dfrac{-2 + 4}{2}, \dfrac{3 + (-5)}{2}\Big) \\[1em] \Rightarrow \Big(\dfrac{2}{2}, \dfrac{-2}{2}\Big) \\[1em] \Rightarrow (1, -1).$

Hence, Option 2 is the correct option.

Question 4

If one end of a diameter of a circle is (2,3) and the centre is (-2, 5), then the other end is

(-6, 7)
(6, -7)
(0, 8)
(0, 4)

Answer

Let B(x, y) be the other end of the diameter, whose one end is A(2, 3)

∴ The mid-point of AB is $\Big(\dfrac{x + 2}{2}, \dfrac{y + 3}{2}\Big)$

The center of the circle is (-2, 5).

Since, the centre of the circle is the mid-point of AB,

$\Rightarrow \dfrac{2 + x}{2} = -2 \text{ and } \dfrac{y + 3}{2} = 5 \\[1em] \Rightarrow 2 + x = -4 \text{ and } y + 3 = 10 \\[1em] \Rightarrow x = -4 - 2 \text{ and } y = 10 - 3 \\[1em] \Rightarrow x = -6 \text{ and } y = 7.$

∴ Coordinates of other end are (-6, 7).

Hence, Option 1 is the correct option.

Question 5

If the mid-point of the line segment joining the points P(a, b - 2) and Q(-2, 4) is R(2, -3), then the values of a and b are

a = 4, b = -5
a = 6, b = 8
a = 6, b = -8
a = -6, b = 8

Answer

The mid-point of the line segment joining the points P(a, b - 2) and Q(-2, 4) is $\dfrac{a + (-2)}{2}, \dfrac{b - 2 + 4}{2}$ .

Given R(2, -3) is the mid-point.

Comparing the values of mid-point we get,

$\dfrac{a + (-2)}{2} = 2 \text{ and } \dfrac{b - 2 + 4}{2} = -3$
⇒ a - 2 = 4 and b + 2 = -6
⇒ a = 4 + 2 and b = -6 - 2
⇒ a = 6 and b = -8.

Hence, Option 3 is the correct option.

Question 6

The point which lies on the perpendicular bisector of the line segment joining the points A(-2, -5) and B(2, 5) is

(0, 0)
(0, 2)
(2, 0)
(-2, 0)

Answer

The point which lies on the perpendicular bisector of the line segment is the mid-point of the line.

Mid-point formula = $\Big(\dfrac{x_1 + x_2}{2} + \dfrac{y_1 + y_2}{2}\Big)$

Putting values to find mid point of AB,

$= \Big(\dfrac{-2 + 2}{2}, \dfrac{-5 + 5}{2}\Big) = (0, 0).$

Hence, Option 1 is the correct option.

Question 7

The coordinates of the point which is equidistant from the three vertices of △AOB (shown in the adjoining figure) are

The coordinates of the point which is equidistant from the three vertices of △AOB(shown in the given figure) are (a) (x, y) (b) (y, x) (c) (x/2, y/2) (d) (y/2, x/2). Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

(x, y)
(y, x)
$\Big(\dfrac{x}{2}, \dfrac{y}{2}\Big)$
$\Big(\dfrac{y}{2}, \dfrac{x}{2}\Big)$

Answer

Since the triangle AOB is the right angled triangle the point which is equidistant from O, A and B is the mid-point of AB. Let that mid-point be D.

Applying mid-point formula, coordinates are,

$= \dfrac{0 + 2x}{2}, \dfrac{2y + 0}{2} \\[1em] = \dfrac{2x}{2}, \dfrac{2y}{2} \\[1em] = (x, y).$

Hence, Option 1 is the correct option.

Question 8

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

(0, 1)
(0, -1)
(-1, 0)
(1, 0)

Answer

ABCD is a parallelogram whose three vertices are A(-2, 3), B(6, 7) and C(8, 3). Let coordinates of its fourth vertex D be (x, y).

The diagonals AC and BD bisect each other at O so O is the mid-point of AC as well as BD.

The fourth vertex D of a parallelogram ABCD whose vertices are A(-2, 3), B(6, 7) and C(8, 3) is. Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

So, by mid-point formula the coordinates of O are,

$= \Big(\dfrac{-2 + 8}{2}, \dfrac{3 + 3}{2}\Big) \\[1em] = \Big(\dfrac{6}{2}, \dfrac{6}{2}\Big) \\[1em] = (3, 3).$

Since, (3, 3) is the mid-point of BD so,

$\Rightarrow 3 = \dfrac{x + 6}{2} \text{ and } 3 = \dfrac{y + 7}{2} \\[1em] \Rightarrow 6 = x + 6 \text{ and } 6 = y + 7 \\[1em] \Rightarrow x = 6 - 6 \text{ and } y = 6 - 7 \\[1em] \Rightarrow x = 0 \text{ and } y = -1.$

∴ Coordinates of D are (0, -1).

Hence, Option 2 is the correct option.

Question 9

The point which divides the line segment joining the points (7, -6) and (3, 4) in the ratio 1 : 2 internally lies in the

Ist quadrant
IInd quadrant
IIIrd quadrant
IVth quadrant

Answer

Let P(x, y) be the point which divides the line segment joining the points.

Given, A(7, -6) and B(3, 4) is divided by P in ratio 1 : 2.

By section formula,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \text{ and } y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{1 \times 3 + 2 \times 7}{1 + 2} \text{ and } \dfrac{1 \times 4 + 2 \times (-6)}{1 + 2} \\[1em] = \dfrac{3 + 14}{3} \text{ and } \dfrac{4 - 12}{3} \\[1em] = \dfrac{17}{3} \text{ and } -\dfrac{8}{3}.$

We see that x is positive and y is negative.

∴ It lies in the fourth quadrant.

Hence, Option 4 is the correct option.

Question 10

The centroid of the triangle whose vertices are (-4, -2), (6, 2) and (4, 6) is

(2, 2)
(2, 3)
(3, 3)
(0, -1)

Answer

By formula,

Centroid of triangle = $\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)$

Centroid of triangle ABC (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 11

A(1, 4), B(4, 1) and C(x, 4) are the vertices of △ABC. If the centroid of the triangles is G(4, 3), then x is equal to

Answer

By formula,

Centroid of triangle = $\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)$

Substituting values we get :

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

Hence, option 3 is the correct option.

Assertion-Reason Type Questions

Question 1

Assertion (A): The point P(3, -1) divides the line segment joining the points A(1, -3) and B(6, 2) internally in the ratio 2 : 3.

Reason (R): The coordinates of the point which divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂ are $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$ .

Assertion (A) is true, Reason (R) is false.
Assertion (A) is false, Reason (R) is true.
Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

By section-formula,

The coordinates of the point which divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂ are

$\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$ .

So, reason (R) is true.

Let us find the point P which divides the line segment joining A(1, -3) and B(6, 2) in the ratio 2 : 3, by section-formula.

$\Rightarrow \text{P} = \Big(\dfrac{2 \times 6 + 3 \times 1}{2 + 3}, \dfrac{2 \times 2 + 3 \times (-3)}{2 + 3}\Big)\\[1em] = \Big(\dfrac{12 + 3}{5}, \dfrac{4 - 9}{5}\Big)\\[1em] = \Big(\dfrac{15}{5}, \dfrac{-5}{5}\Big)\\[1em] = (3, -1)$

So, assertion (A) is true.

Thus, both A and R are correct, and R clearly explains A.

Hence, option 3 is the correct option.

Question 2

Assertion (A): If the coordinates of the mid-points of the sides AB and AC of Δ ABC are D(3, 5) and E(-3, 5) respectively, then BC = 12 units.

Reason (R): The line joining the mid-points of two sides of a triangle is parallel to the third side and equal to half of it.

Assertion (A) is true, Reason (R) is false.
Assertion (A) is false, Reason (R) is true.
Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

By mid-point theorem,

The line joining the mid-points of two sides of a triangle is parallel to the third side and equal to half of it."

So, reason (R) is true.

Given,

D(3, 5) is mid-points of AB and E(-3, 5) is mid-points of AC. By midpoint Theorem:

DE || BC and DE = $\dfrac{1}{2}$ BC.

By distance formula,

Distance between two points = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$DE = \sqrt{((-3) - 3)^2 + (5 - 5)^2}\\[1em] = \sqrt{(-6)^2 + 0^2}\\[1em] = \sqrt{36}\\[1em] = \text{6 units}$

BC = 2 x DE = 2 x 6 = 12 units.

So, assertion (A) is true.

Thus, both A and R are correct, and R clearly explains A.

Hence, option 3 is the correct option.

Question 3

Assertion (A): The point (-5, 0) lies on the x-axis.

Reason (R): The y-coordinates of a point on x-axis is zero.

Assertion (A) is true, Reason (R) is false.
Assertion (A) is false, Reason (R) is true.
Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

We know that,

A point lies on the x-axis if its y-coordinate is 0.

So, reason (R) is true.

Here, the point is (−5, 0), so the y-coordinate is 0.

So, assertion (A) is true.

Thus, both A and R are correct, and R clearly explains A.

Hence, option 3 is the correct option.

Chapter Test

Question 1

The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of the point C are (0, -3). If origin is the mid-point of the base BC, find the coordinates of the points A and B.

Answer

Given, base BC of an equilateral triangle ABC lies on y-axis and coordinates of the point C are (0, -3).

Let coordinates of B be (x, y). Since, origin is the mid-point of the BC. So, by mid-point formula,

$\Rightarrow 0 = \dfrac{x + 0}{2} \text{ and } \dfrac{y - 3}{2} = 0 \\[1em] \Rightarrow \dfrac{x}{2} = 0 \text{ and } y - 3 = 0 \\[1em] \Rightarrow x = 0 \text{ and } y - 3 = 0 \\[1em] \Rightarrow x = 0 \text{ and } y = 3.$

∴ Coordinates of B are (0, 3).

The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of the point C are (0, -3). If origin is the mid-point of the base BC, find the coordinates of the points A and B. Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

From graph we can see that BC = 6 units. Since, ABC is an equilateral triangle so, AB = BC = AC.

Let coordinates of A be (a, 0) as it lies on x-axis.

AB = $\sqrt{(a - 0)^2 + (0 - 3)^2}$

Since AB = 6 units,

$\therefore \sqrt{(a - 0)^2 + (0 - 3)^2} = 6 \\[1em] \Rightarrow \sqrt{a^2 + 9} = 6 \\[1em] \Rightarrow a^2 + 9 = 36 \\[1em] \Rightarrow a^2 = 36 - 9 \\[1em] \Rightarrow a^2 = 27 \\[1em] \Rightarrow a = \sqrt{27} \\[1em] \Rightarrow a = ±3\sqrt{3}.$

∴ Coordinates of A are $(±3\sqrt{3}, 0)$ .

Hence, coordinates of A are $(±3\sqrt{3}, 0)$ and of B are (0, 3).

Question 2

Find the coordinates of the point that divides the line segment joining the points P(5, -2) and Q(9, 6) internally in the ratio of 3 : 1.

Answer

Let R be the point whose co-ordinates are (x, y) which divides PQ in the ratio of 3 : 1.

By section formula, x-coordinate is given by,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 9 + 1 \times 5}{3 + 1} \\[1em] = \dfrac{27 + 5}{4} \\[1em] = \dfrac{32}{4} \\[1em] = 8.$

Similarly y-coordinate is given by,

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 6 + 1 \times (-2)}{3 + 1} \\[1em] = \dfrac{18 + (-2)}{4} \\[1em] = \dfrac{16}{4} \\[1em] = 4.$

∴ R = (8, 4).

Hence, coordinates of point that divides PQ in the ratio 3 : 1 is (8, 4).

Question 3

Find the coordinates of the point P which is three-fourth of the way from A(3, -1) to B(-2, 5).

Answer

Coordinates of A(3, 1) and B(-2, 5).

Let P divides AB in ratio m₁ : m₂. Given P lies on AB such that,

AP = $\dfrac{3}{4}$ AB = $\dfrac{3}{4}$ (AP + PB)
⇒ 4AP = 3AP + 3PB
⇒ 4AP - 3AP = 3PB
⇒ AP = 3PB
⇒ AP : PB = 3 : 1.

∴ m₁ : m₂ = 3 : 1.

Let coordinates of P be (x, y). By section formula,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{(3 \times (-2) + 1 \times 3)}{3 + 1} \\[1em] = \dfrac{-6 + 3}{4} \\[1em] = -\dfrac{3}{4}.$

Similarly applying section formula we get y-coordinate,

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 5 + 1 \times 1}{3 + 1} \\[1em] = \dfrac{15 + 1}{4} \\[1em] = \dfrac{16}{4} \\[1em] = 4.$

∴ P = $\Big(-\dfrac{3}{4}, 4\Big).$

Hence, coordinates of P are $\Big(-\dfrac{3}{4}, 4\Big).$

Question 4

P and Q are the points on the line segment joining the points A(3, -1) and B(-6, 5) such that AP = PQ = QB. Find the coordinates of P and Q.

Answer

Given, AP = PQ = QB

P and Q are the points on the line segment joining the points A(3, -1) and B(-6, 5) such that AP = PQ = QB. Find the coordinates of P and Q. Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

∴ P divides AB in the ratio of 1 : 2 and Q divides it in 2 : 1.

Let coordinates of P be (a, b) and of Q be (c, d)

Applying section formula for x coordinate of P we get,

$a = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{1 \times (-6) + 2 \times 3}{1 + 2} \\[1em] = \dfrac{-6 + 6}{3} \\[1em] = 0.$

Similarly, applying section formula for y coordinate of P we get,

$b = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{1 \times 5 + 2 \times (-1)}{1 + 2} \\[1em] = \dfrac{5 - 2}{3} \\[1em] = \dfrac{3}{3} \\[1em] = 1.$

∴ Coordinates of P = (a, b) = (0, 1).

Applying section formula for x coordinate of Q we get,

$c = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{2 \times (-6) + 1 \times 3}{2 + 1} \\[1em] = \dfrac{-12 + 3}{3} \\[1em] = -3.$

Similarly, applying section formula for y coordinate of Q we get,

$d = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{2 \times 5 + 1 \times (-1)}{2 + 1} \\[1em] = \dfrac{10 + (-1)}{3} \\[1em] = \dfrac{9}{3} \\[1em] = 3.$

∴ Coordinates of Q = (c, d) = (-3, 3).

Hence, coordinates of P = (0, 1) and Q = (-3, 3).

Question 5

The center of a circle is (α + 2, α - 5). Find the value of α, given that the circle passes through points (2, -2) and (8, -2).

Answer

Given O(α + 2, α - 5) is the center of the circle. A and B are the points on the circle. So, we can say OA = OB = radius.

From distance formula we get,

$OA = \sqrt{(2 - (α + 2))^2 + (-2 - (α - 5))^2} \\[1em] = \sqrt{(2 - α - 2)^2 + (-2 - α + 5)^2} \\[1em] = \sqrt{(-α)^2 + (-α + 3)^2} \\[1em] = \sqrt{α^2 + α^2 + 9 - 6α} \\[1em] = \sqrt{2α^2 - 6α + 9} \qquad \text{....[Eq 1]} \\[1em] OB = \sqrt{(8 - (α + 2))^2 + (-2 - (α - 5))^2} \\[1em] = \sqrt{(8 - α - 2)^2 + (-2 - α + 5)^2} \\[1em] = \sqrt{(6 - α)^2 + (3 - α)^2} \\[1em] = \sqrt{36 + α^2 - 12α + 9 + α^2 - 6α} \\[1em] = \sqrt{2α^2 + 45 - 18α} \qquad \text{....[Eq 2]}$

Comparing both the Equation, since they are equal to radius,

$\Rightarrow \sqrt{2α^2 - 6α + 9} = \sqrt{2α^2 + 45 - 18α} \\[1em]$

Squaring both sides we get,

$\Rightarrow 2α^2 - 6α + 9 = 2α^2 + 45 - 18α \\[1em] \Rightarrow 2α^2 - 2α^2 - 6α + 18α + 9 - 45 = 0 \\[1em] \Rightarrow 12α - 36 = 0 \\[1em] \Rightarrow 12α = 36 \\[1em] \Rightarrow α = 3.$

Hence, the value of α = 3.

Question 6

The mid-point of the line segment joining A(2, p) and B(q, 4) is (3, 5). Calculate the values of p and q.

Answer

Given, (3, 5) is the mid-point of A(2, p) and B(q, 4).

By mid-point formula,

$\Rightarrow 3 = \dfrac{(2 + q)}{2} \text{ and } 5 = \dfrac{(p + 4)}{2} \\[1em] \Rightarrow 2 + q = 6 \text{ and } p + 4 = 10 \\[1em] \Rightarrow q = 4 \text{ and } p = 6.$

Hence, p = 6 and q = 4.

Question 7

The ends of a diameter of a circle have the coordinates (3, 0) and (-5, 6). PQ is another diameter where Q has the coordinates (-1, -2). Find the coordinates of P and the radius of the circle.

Answer

Let AB be the diameter where coordinates of A are (3, 0) and of B are (-5, 6).

∴ Coordinates of its midpoint will be $\Big(\dfrac{3 + (-5)}{2}, \dfrac{0 + 6}{2}\Big)$ or (-1, 3).

Now PQ is another diameter in which the coordinates of Q are (-1, -2).

Let coordinates of P be (x, y), then by mid-point formula coordinates of mid-point will be $\Big(\dfrac{-1 + x}{2}, \dfrac{-2 + y}{2}\Big)$

Since, the diameters of circle intersect at their midpoint.

$\therefore \dfrac{-1 + x}{2} = -1 \text{ and } \dfrac{-2 + y}{2} = 3 \\[1em] \Rightarrow -1 + x = -2 \text{ and } -2 + y = 6 \\[1em] \Rightarrow x = -2 + 1 \text{ and } y = 6 + 2 \\[1em] \Rightarrow x = -1 \text{ and } y = 8.$

∴ Coordinates of P will be (-1, 8).

Radius = OP, by distance-formula we get,

$OP = \sqrt{(-1 - (-1))^2 + (8 - 3)^2} \\[1em] = \sqrt{(-1 + 1)^2 + (5)^2} \\[1em] = \sqrt{0^2 + 25} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.$

Hence, the radius of circle is 5 units and the coordinates of P is (-1, 8).

Question 8

In what ratio does the point (-4, 6) divide the line segment joining the points A(-6, 10) and B(3, -8) ?

Answer

Let the point (-4, 6) divide the line segment joining the points A(-6, 10) and B(3, -8) in the ratio m : n

Using section-formula,

x-coordinate = $\Big(\dfrac{mx_2 + nx_1}{m + n}\Big)$

Comparing,

$\Rightarrow -4 = \Big(\dfrac{m \times 3 + n \times (-6)}{m + n} \Big) \\[1em] \Rightarrow -4 = \dfrac{3m - 6n}{m + n} \\[1em] \Rightarrow -4(m + n) = 3m - 6n \\[1em] \Rightarrow -4m -4n = 3m - 6n \\[1em] \Rightarrow -4m - 3m = -6n + 4n \\[1em] \Rightarrow -7m = -2n \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{2}{7} \\[1em] \Rightarrow m : n = 2 : 7.$

The ratio in which the point (-4, 6) divides the line segment is 2 : 7.

Question 9

Find the ratio in which the point P(-3, p) divides the line segment joining the points (-5, -4) and (-2, 3). Hence, find the value of p.

Answer

Let (-3, p) divides the line segment in the ratio of m : n.

By section-formula,

x-coordinate = $\Big(\dfrac{mx_2 + nx_1}{m + n}\Big)$

$\therefore -3 = \dfrac{m \times (-2) + n \times (-5)}{m + n} \\[1em] \Rightarrow -3 = \dfrac{-2m - 5n}{m + n} \\[1em] \Rightarrow -3(m + n) = -2m - 5n \\[1em] \Rightarrow -3m - 3n = -2m - 5n \\[1em] \Rightarrow -3m + 2m = -5n + 3n \\[1em] \Rightarrow -m = -2n \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{2}{1} \\[1em] \Rightarrow m : n = 2 : 1. \\[1em]$

By section formula, y-coordinate = p

$= \dfrac{my_2 + ny_1}{m + n} \\[1em] = \dfrac{2 \times 3 + 1 \times (-4)}{2 + 1} \\[1em] = \dfrac{6 - 4}{3} \\[1em] = \dfrac{2}{3}.$

Hence, the value of p = $\dfrac{2}{3}$ and the ratio in which point P divides the line segment is 2 : 1.

Question 10

In what ratio is the line joining the points (4, 2) and (3, -5) divided by the x-axis? Also find the coordinates of the point of division.

Answer

Let the point P which is on the x-axis, divide the line segment joining the points A(4, 2) and B(3, -5) in the ratio of m : n. Let the coordinates of P be (x, 0).

By section formula,

y-coordinate = $\dfrac{my_2 + ny_1}{m + n}$

$\Rightarrow 0 = \dfrac{m \times (-5) + n \times 2}{m + n} \\[1em] \Rightarrow 0 = -5m + 2n \\[1em] \Rightarrow 5m = 2n \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{2}{5} \\[1em] \Rightarrow m : n = 2 : 5.$

Putting value of m : n in section-formula for x-coordinate,

x-coordinate = $\dfrac{mx_2 + nx_1}{m+ n}$

$= \dfrac{2 \times 3 + 5 \times 4}{2 + 5} \\[1em] = \dfrac{6 + 20}{7} \\[1em] = \dfrac{26}{7}.$

Hence, coordinates of P are $(\dfrac{26}{7}, 0)$ and 2 : 5 is the ratio in which the line joining the points (4, 2) and (3, -5) is divided by the x-axis.

Question 11

If the abscissa of a point P is 2, find the ratio in which it divides the line segment joining the points (-4, 3) and (6, 3). Hence, find the coordinates of P.

Answer

Let coordinates of A be (-4, 3) and of B be (6, 3) and of P be (2, y).

Let the ratio in which the P divides AB be m : n.

If the abscissa of a point P is 2, find the ratio in which it divides the line segment joining the points (-4, 3) and (6, 3). Hence, find the coordinates of P. Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

By section formula,

x-coordinate = $\dfrac{mx_2 + nx_1}{m+ n}$

$\Rightarrow 2 = \dfrac{m \times 6 + n \times (-4)}{m + n} \\[1em] \Rightarrow 2 = \dfrac{6m - 4n}{m + n} \\[1em] \Rightarrow 2(m + n) = 6m - 4n \\[1em] \Rightarrow 2m + 2n = 6m - 4n \\[1em] \Rightarrow 2n + 4n = 6m - 2m \\[1em] \Rightarrow 6n = 4m \\[1em] \Rightarrow 6n = 4m \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{6}{4} \\[1em] \Rightarrow m : n = 3 : 2.$

Similarly for y-coordinate,

$\Rightarrow y = \dfrac{my_2 + ny_1}{m + n} \\[1em] = \dfrac{3 \times 3 + 2 \times 3}{3 + 2} \\[1em] = \dfrac{9 + 6}{5} \\[1em] = \dfrac{15}{5} \\[1em] = 3.$

Hence, the coordinates of P is (2, 3) and it divides the line in the ratio 3 : 2.

Question 12

Determine the ratio in which the line 2x + y - 4 = 0 divide the line segment joining the points A(2, -2) and B(3, 7). Also find the coordinates of the point of the division.

Answer

Let the line 2x + y - 4 divide the line segment AB in the ratio m : n at P. So, by section-formula coordinates of P are,

x-coordinate = x = $\dfrac{m \times x_2 + n \times x_1}{m + n}$

$= \dfrac{m \times 3 + n \times 2}{m + n} \\[1em] = \dfrac{3m + 2n}{m + n} \qquad \text{....[Eq 1]} \\[1em]$

Similarly,

y-coordinate = y = $\dfrac{m \times y_2 + n \times y_1}{m + n}$

$= \dfrac{m \times 7 + n \times (-2)}{m + n} \\[1em] = \dfrac{7m - 2n}{m + n} \qquad \text{....[Eq 2]} \\[1em]$

Since, P lies on the line 2x + y - 4 = 0.

$\therefore 2\dfrac{(3m + 2n)}{m + n} + \dfrac{7m - 2n}{m + n} - 4 = 0 \\[1em] \Rightarrow \dfrac{6m + 4n + 7m - 2n -4(m + n)}{m + n} = 0 \\[1em] \Rightarrow 6m + 4n + 7m - 2n - 4m - 4n = 0 \\[1em] \Rightarrow 9m - 2n = 0 \\[1em] \Rightarrow 9m = 2n \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{2}{9} \\[1em]$

Putting values in Eq 1 for x-coordinate we get,

$\therefore x = \dfrac{3 \times 2 + 2 \times 9}{2 + 9} \\[1em] = \dfrac{6 + 18}{11} \\[1em] = \dfrac{24}{11}. \\[1em]$

Putting values in Eq 2 for y-coordinate we get,

$\text{y = } \dfrac{7 \times 2 - 2 \times 9}{2 + 9} \\[1em] = \dfrac{14 - 18}{11} \\[1em] = -\dfrac{4}{11}.$

Hence, coordinates of P will be $\Big(\dfrac{24}{11}, -\dfrac{4}{11}\Big)$ and 2 : 9 is the ratio in which the line 2x + y - 4 divides AB.

Question 13

ABCD is a parallelogram. If the coordinates of A, B and D are (10, -6), (2, -6) and (4, -2) respectively, find the coordinates of C.

Answer

Let the coordinates of C be (x, y) and other three vertices of the given parallelogram are A(10, -6), B(2, -6) and D(4, -2).

Since, ABCD is a parallelogram, its diagonals bisect each other.

ABCD is a parallelogram. If the coordinates of A, B and D are (10, -6), (2, -6) and (4, -2) respectively, find the coordinates of C. Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Let AC and BD intersect each other at O.

So, O is the mid-point of BD, so coordinates of O are,

$\Rightarrow \Big(\dfrac{2 + 4}{2}, \dfrac{-6 + (-2)}{2}\Big) \\[1em] = \Big(\dfrac{6}{2}, \dfrac{-8}{2}\Big) \\[1em] = (3, -4).$

The mid-point of AC is

$\Rightarrow \Big(\dfrac{10 + x}{2}, \dfrac{-6 + y}{2}\Big)$

Since O is the mid-point of AC so comparing,

$\Rightarrow \dfrac{10 + x}{2} = 3 \text{ and } \dfrac{-6 + y}{2} = -4 \\[1em] \Rightarrow 10 + x = 6 \text{ and } -6 + y = -8 \\[1em] \Rightarrow x = 6 - 10 \text{ and } y = -8 + 6 \\[1em] \Rightarrow x = -4 \text{ and } y = -2.$

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

Question 14

ABCD is a parallelogram whose vertices A and B have coordinates (2, -3) and (-1, -1) respectively. If the diagonals of the parallelogram meet at the point M(1, -4), find the coordinates of C and D. Hence, find the perimeter of the parallelogram.

Answer

Coordinates of A are (2, -3) and B (-1, -1).

Since, M is the point where diagonals meet, hence it is midpoint of AC and BD.

Let coordinates of C and D be (x₁, y₁) and (x₂, y₂).

ABCD is a parallelogram whose vertices A and B have coordinates (2, -3) and (-1, -1) respectively. If the diagonals of the parallelogram meet at the point M(1, -4), find the coordinates of C and D. Hence, find the perimeter of the parallelogram. Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

When M(1, -4) is the midpoint of AC then by midpoint formula,

$\Rightarrow 1 = \dfrac{2 + x_1}{2} \text{ and } -4 = \dfrac{-3 + y_1}{2} \\[1em] \Rightarrow 2 + x_1 = 2 \text{ and } -3 + y_1 = -8 \\[1em] \Rightarrow x_1 = 2 - 2 \text{ and } y_1 = -8 + 3 \\[1em] \Rightarrow x_1 = 0 \text{ and } y_1 = -5.$

∴ Coordinates of C are (0, -5)

When M(1, -4) is the midpoint of BD then by midpoint formula,

$\Rightarrow 1 = \dfrac{-1 + x_2}{2} \text{ and } -4 = \dfrac{-1 + y_2}{2} \\[1em] \Rightarrow -1 + x_2 = 2 \text{ and } -1 + y_2 = -8 \\[1em] \Rightarrow x_2 = 2 + 1 \text{ and } y_2 = -8 + 1 \\[1em] \Rightarrow x_2 = 3 \text{ and } y_2 = -7.$

∴ Coordinates of D are (3, -7).

By distance formula, the length of AB is,

$= \sqrt{[2 - (-1)]^2 + [(-3) - (-1)]^2} \\[1em] = \sqrt{(2 + 1)^2 + (-3 + 1)^2} \\[1em] = \sqrt{3^2 + (-2)^2} \\[1em] = \sqrt{9 + 4} \\[1em] = \sqrt{13}.$

By distance formula, the length of BC is,

$= \sqrt{[0 - (-1)]^2 + [(-5) - (-1)]^2} \\[1em] = \sqrt{(1)^2 + (-5 + 1)^2} \\[1em] = \sqrt{1 + (-4)^2} \\[1em] = \sqrt{1 + 16} \\[1em] = \sqrt{17}.$

Perimeter of parallelogram ABCD = 2(AB + BC) = $2(\sqrt{13} + \sqrt{17}).$

Hence, the coordinates of C and D are (0, -5) and (3, -7) respectively. The perimeter of parallelogram ABCD is $2(\sqrt{13} + \sqrt{17})$ units.

Question 15

Given O (0, 0), P(1, 2), S(-3, 0). P divides OQ in the ratio 2 : 3 and OPRS is a parallelogram.

Find :

(i) the coordinates of Q.

(ii) the coordinates of R.

(iii) the ratio in which RQ is divided by the x-axis.

Answer

(i) Let coordinates of Q be (a, b).

Given, point P(1, 2) divides OQ in the ratio of 2 : 3. Here, O(0, 0) is the origin.

By section formula we get x,

x-coordinate = $\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}$

$\Rightarrow 1 = \dfrac{2 \times a + 3 \times 0}{2 + 3} \\[1em] \Rightarrow 1 = \dfrac{2a}{5} \\[1em] \Rightarrow a = \dfrac{5}{2}.$

By section formula we get y,

y-coordinate = $\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}$

$\Rightarrow 2 = \dfrac{2 \times b + 3 \times 0}{2 + 3} \\[1em] \Rightarrow 2 = \dfrac{2b}{5} \\[1em] \Rightarrow b = 5.$

Hence, coordinates of Q are $(\dfrac{5}{2}, 5)$ .

(ii) In OPRS, OR and PS are diagonals. Let them bisect each other at point M which is the mid-point of both the diagonals. Let coordinates of R be (c, d).

Since, M is the midpoint of PS, by mid-point formula, coordinates of M

$= \Big(\dfrac{1 + (-3)}{2}, \dfrac{2 + 0}{2}\Big) \\[1em] = \Big(\dfrac{-2}{2}, \dfrac{2}{2}\Big) \\[1em] = (-1, 1).$

Since, M is the mid-point of OR also so,

$\Rightarrow -1 = \dfrac{0 + c}{2} \text{ and } 1 = \dfrac{0 + d}{2} \\[1em] \Rightarrow c = -2 \text{ and } d = 2.$

Hence, coordinates of R are (-2, 2).

(iii) Let the point on y-axis that divides RQ is N and it divides in ratio m₁ : m₂.

Since, N lies on y-axis so abscissa(x) = 0.

By section formula we get,

$\text{x-coordinate } = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 0 = \dfrac{m_1 \times \dfrac{5}{2} + m_2 \times (-2)}{m_1 + m_2} \\[1em] \Rightarrow \dfrac{5m_1}{2} - 2m_2 = 0 \\[1em] \Rightarrow \dfrac{5}{2}m_1 = 2m_2 \\[1em] \Rightarrow m_1 : m_2 = \dfrac{2 \times 2}{5} \\[1em] \Rightarrow m_1 : m_2 = 4 : 5.$

Hence, RQ is divided in the ratio 4 : 5 by x-axis.

Question 16

If A(5, -1), B(-3, -2) and C(-1, 8) are the vertices of a triangle ABC, find the length of the median through A and the coordinates of the centroid of triangle ABC.

Answer

A(5, -1), B(-3, -2) and C(-1, 8) are the vertices of △ABC. Below figure shows the triangle:

If A(5, -1), B(-3, -2) and C(-1, 8) are the vertices of a triangle ABC, find the length of the median through A and the coordinates of the centroid of triangle ABC. Section Formula, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Let D be the midpoint of BC. By midpoint formula, coordinates of D are,

$= \Big(\dfrac{-3 + (-1)}{2}, \dfrac{-2 + 8}{2}\Big) \\[1em] = \Big(\dfrac{-4}{2}, \dfrac{6}{2}\Big) \\[1em] = (-2, 3).$

By distance formula we get,

$AD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] = \sqrt{(-2 - 5)^2 + (3 - (-1))^2} \\[1em] = \sqrt{(-7)^2 + (3 + 1)^2} \\[1em] = \sqrt{49 + 16} \\[1em] = \sqrt{65} \text{ units.}$

Coordinates of centroid is given by $\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)$

$= \Big(\dfrac{5 + (-3) + (-1)}{3}, \dfrac{-1 + (-2) + 8}{3}\Big) \\[1em] = \Big(\dfrac{1}{3}, \dfrac{5}{3}\Big).$

Hence, the coordinates of the centroid of triangle is $\Big(\dfrac{1}{3}, \dfrac{5}{3}\Big)$ and the length of the median through A is $\sqrt{65}$ units.

Chapter 11

Section Formula

Class - 10 ML Aggarwal Understanding ICSE Mathematics

Exercise 11

Question 1

Question 2

Question 3(i)

Question 3(ii)

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13(i)

Question 13(ii)

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question 24

Question 25

Question 26

Question 27

Question 28

Question 29

Multiple Choice Questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Assertion-Reason Type Questions

Question 1

Question 2

Question 3

Chapter Test

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16