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Chapter 25

Distance Formula

Class - 9 Concise Mathematics Selina

Exercise 25

Question 1(a)

The distance between the points (7, -5) and (3, -1) is :

  1. 4 units

  2. 3 units

  3. 424{\sqrt2} units

  4. 5 units

Answer

Let (7, -5) = (x1, y1) and (3, -1) = (x2, y2)

Distance between the given points =

(x2−x1)2+(y2−y1)2=(3−7)2+((−1)−(−5))2=(−4)2+(4)2=16+16=32=42units\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{(3 - 7)^2 + ((-1) - (-5))^2}\\[1em] = \sqrt{(-4)^2 + (4)^2}\\[1em] = \sqrt{16 + 16}\\[1em] = \sqrt{32}\\[1em] = 4{\sqrt2} \text{units}

Hence, option 3 is the correct option.

Question 1(b)

The distance of point (-4, 3) from the origin is :

  1. 5 units

  2. -5 units

  3. 4 units

  4. 3 units

Answer

Since, distance between origin and (x, y) = x2+y2\sqrt{x^2 + y^2}

∴ Distance between origin and the point (-4, 3)

=(−4)2+32=16+9=25=5 units= \sqrt{(-4)^2 + 3^2}\\[1em] = \sqrt{16 + 9}\\[1em] = \sqrt{25}\\[1em] = \text{5 units}

Hence, option 1 is the correct option.

Question 1(c)

The distance between the points (-3, 2) and (x, 10) is 10 units. The value of x is :

  1. 3

  2. -9

  3. 3 or -9

  4. 3 and -9

Answer

Let (-3, 2) = (x1, y1) and (x, 10) = (x2, y2)

Distance between the given points =

(x2−x1)2+(y2−y1)210=(x−(−3))2+(10−2)2⇒10=(x+3)2+82⇒102=(x+3)2+(8)2⇒100=(x+3)2+64⇒100−64=(x+3)2⇒36=(3+x)2⇒36=3+x±6=3+x6=x+3 or−6=3+xx=6−3 or−6−3=xx=3 or−9\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] 10 = \sqrt{(x - (-3))^2 + (10 - 2)^2}\\[1em] ⇒ 10 = \sqrt{(x + 3)^2 + 8^2}\\[1em] ⇒ 10^2 = (x + 3)^2 + (8)^2\\[1em] ⇒ 100 = (x + 3)^2 + 64\\[1em] ⇒ 100 - 64 = (x + 3)^2\\[1em] ⇒ 36 = (3 + x)^2\\[1em] ⇒ \sqrt{36} = 3 + x\\[1em] ± 6 = 3 + x\\[1em] 6 = x + 3 \text{ or} -6 = 3 + x \\[1em] x = 6 - 3 \text{ or} -6 -3 = x \\[1em] x = 3 \text{ or} -9\\[1em]

Hence, option 3 is the correct option.

Question 1(d)

The point (x, y) is equidistant from the points (3, 6) and (-3, 4); the relation between x and y is :

  1. 3x - y = 5

  2. 3x - y - 5 = 0

  3. y - 3x = 0

  4. 3x + y = 5

Answer

Given (x, y) is equidistant from (3, 6) and (-3, 4)

i.e. distance between (x, y) and (3, 6) = distance between (x, y) and (-3, 4)

(x−3)2+(y−6)2=(x−(−3))2+(y−4)2⇒(x−3)2+(y−6)2=(x+3)2+(y−4)2⇒(x−3)2+(y−6)2=(x+3)2+(y−4)2⇒x2+9−6x+y2+36−12y=x2+9+6x+y2+16−8y⇒36−16=6x+6x+12y−8y⇒20=12x+4y⇒3x+y=5\sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x - (-3))^2 + (y - 4)^2}\\[1em] ⇒ \sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x + 3)^2 + (y - 4)^2}\\[1em] ⇒ (x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2\\[1em] ⇒ x^2 + 9 - 6x + y^2 + 36 - 12y = x^2 + 9 + 6x + y^2 + 16 -8y\\[1em] ⇒ 36 - 16 = 6x + 6x + 12y -8y\\[1em] ⇒ 20 = 12x + 4y\\[1em] ⇒ 3x + y = 5

Hence, option 4 is the correct option.

Question 1(e)

The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is :

  1. 12 units

  2. 6 units

  3. 5 units

  4. 10 units

Answer

Let (0, 4) = (x1, y1), (0, 0) = (x2, y2) and (3, 0) = (x3, y3)

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

The perimeter of a triangle

=(x2−x1)2+(y2−y1)2+(x3−x2)2+(y3−y2)2+(x3−x1)2+(y3−y1)2=(0−0)2+(0−4)2+(3−0)2+(0−0)2+(3−0)2+(0−4)2=(−4)2+(3)2+(3)2+(−4)2=16+9+9+16=4+3+25=4+3+5=12units= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} + \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2} + \sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2}\\[1em] = \sqrt{(0 - 0)^2 + (0 - 4)^2} + \sqrt{(3 - 0)^2 + (0 - 0)^2} + \sqrt{(3 - 0)^2 + (0 - 4)^2}\\[1em] = \sqrt{(- 4)^2} + \sqrt{(3)^2} + \sqrt{(3)^2 + (- 4)^2}\\[1em] = \sqrt{16} + \sqrt{9} + \sqrt{9 + 16}\\[1em] = 4 + 3 + \sqrt{25}\\[1em] = 4 + 3 + 5\\[1em] = 12 \text{units}

Hence, option 1 is the correct option.

Question 2

Find the distance between the following pairs of points :

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

(ii) (-a, -b) and (a, b)

(iii) (35,2)\Big(\dfrac{3}{5},2\Big) and (−15,125)\Big(-\dfrac{1}{5}, 1\dfrac{2}{5}\Big)

(iv) (3+1,1)\Big({\sqrt3 +1},1\Big) and (0, 3\sqrt{3})

Answer

(i) Let (-3, 6) = (x1, y1) and (2, -6) = (x2, y2)

⇒ Distance between the given points =

(x2−x1)2+(y2−y1)2=(2−(−3))2+(−6−6)2=(5)2+(−12)2=25+144=169=13\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{(2 - (-3))^2 + (-6 - 6)^2}\\[1em] = \sqrt{(5)^2 + (-12)^2}\\[1em] = \sqrt{25 + 144}\\[1em] = \sqrt{169}\\[1em] = 13

Hence, distance between the given points is 13.

(ii) Let (-a, -b) = (x1, y1) and (a, b) = (x2, y2)

⇒ Distance between the given points =

(x2−x1)2+(y2−y1)2=(a−(−a))2+(b−(−b))2=(2a)2+(2b)2=4a2+4b2=2a2+b2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{(a - (-a))^2 + (b - (-b))^2}\\[1em] = \sqrt{(2a)^2 + (2b)^2}\\[1em] = \sqrt{4a^2 + 4b^2}\\[1em] = 2\sqrt{a^2 + b^2}\\[1em]

Hence, distance between the given points is 2a2+b22\sqrt{a^2 + b^2}.

(iii) Let (35,2)\Big(\dfrac{3}{5},2\Big) = (x1, y1) and (−15,125)\Big(-\dfrac{1}{5}, 1\dfrac{2}{5}\Big) = (x2, y2)

⇒ Distance between the given points =

(x2−x1)2+(y2−y1)2=(−15−35)2+((125−2))2=(−45)2+(7−105)2=1625+(−35)2=1625+925=2525=1=1\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{\Big(\dfrac{-1}{5} - \dfrac{3}{5} \Big)^2 + \Big(\Big(1\dfrac{2}{5} - 2 \Big)\Big)^2}\\[1em] = \sqrt{\Big(\dfrac{-4}{5}\Big)^2 + \Big(\dfrac{7-10}{5}\Big)^2}\\[1em] = \sqrt{\dfrac{16}{25} + \Big(\dfrac{-3}{5}\Big)^2}\\[1em] = \sqrt{\dfrac{16}{25} + \dfrac{9}{25}}\\[1em] = \sqrt{\dfrac{25}{25}}\\[1em] = \sqrt{1}\\[1em] = 1

Hence, distance between the given points is 1.

(iv) Let (3+1,1)\Big({\sqrt3 +1},1\Big) = (x1, y1) and (0, 3\sqrt{3}) = (x2, y2)

⇒ Distance between the given points =

(x2−x1)2+(y2−y1)2=(0−(3+1))2+(3−1)2=(−(3+1))2+(3−1)2=3+1+23+1+3−23=8=22\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{(0 - ({\sqrt3 +1}))^2 + (\sqrt3 - 1)^2}\\[1em] = \sqrt{(-{(\sqrt3 +1)})^2 + (\sqrt3 - 1)^2}\\[1em] = \sqrt{3 + 1 + 2\sqrt3 + 1 + 3 - 2\sqrt3}\\[1em] = \sqrt{8}\\[1em] = 2\sqrt2

Hence, distance between the given points is 222\sqrt2 = 2.83.

Question 3

Find the distance between the origin and the point :

(i) (-8, 6)

(ii) (-5, -12)

(iii) (8, -15)

Answer

(i) Since, distance between origin and (x, y) = x2+y2\sqrt{x^2 + y^2}

∴ Distance between origin and the point (-8, 6)

=(−8)2+62=64+36=100=10= \sqrt{(-8)^2 + 6^2}\\[1em] = \sqrt{64 + 36}\\[1em] = \sqrt{100}\\[1em] = \text{10}

Hence, the distance between the origin and the point (-8, 6) is 10.

(ii) Since, distance between origin and (x, y) = x2+y2\sqrt{x^2 + y^2}

∴ Distance between origin and the point (-5, -12)

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

Hence, the distance between the origin and the point (-5, -12) is 13.

(iii) Since, distance between origin and (x, y) = x2+y2\sqrt{x^2 + y^2}

∴ Distance between origin and the point (8, -15)

=82+(−15)2=64+225=289=17= \sqrt{8^2 + (-15)^2}\\[1em] = \sqrt{64 + 225}\\[1em] = \sqrt{289}\\[1em] = \text{17}

Hence, the distance between the origin and the point is 17.

Question 4

The distance between the points (3, 1) and (0, x) is 5. Find x.

Answer

Let (3, 1) = (x1, y1) and (0, x) = (x2, y2)

⇒ Distance between the given points =

(x2−x1)2+(y2−y1)2⇒5=(0−3)2+(x−1)2⇒5=(−3)2+(x−1)2⇒52=(−3)2+(x−1)2⇒25=9+(x−1)2⇒25−9=(x−1)2⇒16=(x−1)2⇒16=x−1⇒4=x−1 and−4=x−1⇒x=4+1 and x=−4+1⇒x=5 and−3\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] ⇒ 5 = \sqrt{(0 - 3)^2 + (x - 1)^2}\\[1em] ⇒ 5 = \sqrt{(-3)^2 + (x - 1)^2}\\[1em] ⇒ 5^2 = (-3)^2 + (x - 1)^2\\[1em] ⇒ 25 = 9 + (x - 1)^2\\[1em] ⇒ 25 - 9 = (x - 1)^2\\[1em] ⇒ 16 = (x - 1)^2\\[1em] ⇒ \sqrt{16} = x - 1\\[1em] ⇒ 4 = x - 1 \text{ and} -4 = x -1\\[1em] ⇒ x = 4 + 1 \text{ and } x = -4 + 1\\[1em] ⇒ x = 5 \text{ and} -3\\[1em]

Hence, the value of x = 5 and -3.

Question 5

Find the co-ordinates of points on the x-axis which are at a distance of 17 units from the point (11, -8).

Answer

Let the co-ordinates of the point on the x-axis be (x, 0).

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

Let (x, 0) = (x1, y1) and (11, -8) = (x2, y2)

⇒ Distance between the given points =

⇒17=(11−x)2+(−8−0)2⇒17=(11−x)2+(−8)2⇒172=(11−x)2+(−8)2⇒289=121+x2−22x+64⇒289=185+x2−22x⇒185+x2−22x−289=0⇒x2−(26x−4x)−104=0⇒x2−26x+4x−104=0⇒(x2−26x)+(4x−104)=0⇒x(x−26)+4(x−26)=0⇒(x−26)(x+4)=0⇒x=26 and−4\\[1em] ⇒ 17 = \sqrt{(11 - x)^2 + (-8 - 0)^2}\\[1em] ⇒ 17 = \sqrt{(11 - x)^2 + (-8)^2}\\[1em] ⇒ 17^2 = (11 - x)^2 + (-8)^2\\[1em] ⇒ 289 = 121 + x^2 - 22x + 64\\[1em] ⇒ 289 = 185 + x^2 - 22x\\[1em] ⇒ 185 + x^2 - 22x - 289 = 0\\[1em] ⇒ x^2 - (26x - 4x) - 104 = 0\\[1em] ⇒ x^2 - 26x + 4x - 104 = 0\\[1em] ⇒ (x^2 - 26x) + (4x - 104) = 0\\[1em] ⇒ x(x - 26) + 4(x - 26) = 0\\[1em] ⇒ (x - 26)(x + 4) = 0\\[1em] ⇒ x = 26 \text{ and} -4

Hence, the co-ordinates of points are (26, 0) and (-4, 0).

Question 6

Find the co-ordinates of the points on the y-axis, which are at a distance of 10 units from the point (-8, 4).

Answer

Let (0, y) = (x1, y1) and (-8, 4) = (x2, y2)

⇒ Distance between the given points =

(x2−x1)2+(y2−y1)2⇒10=(−8−0)2+(4−y)2⇒10=(−8)2+(4−y)2⇒102=64+16+y2−8y⇒100=80+y2−8y⇒80+y2−8y−100=0⇒y2−8y−20=0⇒y2−(10y−2y)−20=0⇒y2−10y+2y−20=0⇒(y2−10y)+(2y−20)=0⇒y(y−10)+2(y−10)=0⇒(y−10)(y+2)=0⇒y=10 and −2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] ⇒ 10 = \sqrt{(-8 - 0)^2 + (4 - y)^2}\\[1em] ⇒ 10 = \sqrt{(-8)^2 + (4 - y)^2}\\[1em] ⇒ 10^2 = 64 + 16 + y^2 - 8y\\[1em] ⇒ 100 = 80 + y^2 - 8y\\[1em] ⇒ 80 + y^2 - 8y - 100 = 0 \\[1em] ⇒ y^2 - 8y - 20 = 0 \\[1em] ⇒ y^2 - (10y - 2y) - 20 = 0 \\[1em] ⇒ y^2 - 10y + 2y - 20 = 0 \\[1em] ⇒ (y^2 - 10y) + (2y - 20) = 0 \\[1em] ⇒ y(y - 10) + 2(y - 10) = 0 \\[1em] ⇒ (y - 10)(y + 2) = 0 \\[1em] ⇒ y = 10 \text{ and } -2

Hence, the co-ordinates of points are (0, 10) and (0, -2).

Question 7

A point A is at a distance of 10\sqrt{10} units from the point (4, 3). Find the co-ordinates of point A, if its ordinate is twice its abscissa.

Answer

Let (a, 2a) = (x1, y1) and (4, 3) = (x2, y2)

⇒ Distance between the given points =

(x2−x1)2+(y2−y1)2⇒10=(4−a)2+(3−2a)2⇒10=(4−a)2+(3−2a)2⇒10=16+a2−8a+9+4a2−12a⇒10=25+5a2−20a⇒25+5a2−20a−10=0⇒5a2−20a+15=0⇒a2−4a+3=0⇒a2−(3a+1a)+3=0⇒a2−3a−1a+3=0⇒a(a−3)−1(a−3)=0⇒(a−3)(a−1)=0⇒a=3 and 1\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] ⇒ \sqrt{10} = \sqrt{(4 - a)^2 + (3 - 2a)^2}\\[1em] ⇒ 10 = (4 - a)^2 + (3 - 2a)^2\\[1em] ⇒ 10 = 16 + a^2 - 8a + 9 + 4a^2 - 12a\\[1em] ⇒ 10 = 25 + 5a^2 - 20a\\[1em] ⇒ 25 + 5a^2 - 20a - 10 = 0\\[1em] ⇒ 5a^2 - 20a + 15 = 0\\[1em] ⇒ a^2 - 4a + 3 = 0\\[1em] ⇒ a^2 - (3a + 1a) + 3 = 0\\[1em] ⇒ a^2 - 3a - 1a + 3 = 0\\[1em] ⇒ a(a - 3) - 1(a - 3) = 0\\[1em] ⇒ (a - 3)(a - 1) = 0\\[1em] ⇒ a = 3 \text{ and } 1

For each value of a = x, we can find the corresponding value of y:

  • If a = 3, then y = 2a = 6
  • If a = 1, then y = 2a = 2

Hence, the co-ordinates of point A are (3, 6) and (1, 2).

Question 8

A point P (2, -1) is equidistant from the points (a, 7) and (-3, a). Find a.

Answer

Given point (2, -1) is equidistant from (a, 7) and (-3, a).

i.e. distance between (2, -1) and (a, 7) = distance between (2, -1) and (-3, a)

(a−2)2+(7−(−1))2=((−3)−2)2+(a−(−1))2⇒(a−2)2+(7+1)2=(−3−2)2+(a+1)2⇒(a−2)2+(8)2=(−5)2+(a+1)2⇒(a−2)2+(8)2=(−5)2+(a+1)2⇒a2+4−4a+64=25+a2+1+2a⇒a2−4a+68=a2+2a+26⇒−4a+68=2a+26⇒68−26=2a+4a⇒6a=42⇒a=426⇒a=7\sqrt{(a - 2)^2 + (7 - (-1))^2} = \sqrt{((-3) - 2)^2 + (a - (-1))^2}\\[1em] ⇒ \sqrt{(a - 2)^2 + (7 + 1)^2} = \sqrt{(-3 - 2)^2 + (a + 1)^2}\\[1em] ⇒ \sqrt{(a - 2)^2 + (8)^2} = \sqrt{(-5)^2 + (a + 1)^2}\\[1em] ⇒ (a - 2)^2 + (8)^2 = (-5)^2 + (a + 1)^2\\[1em] ⇒ a^2 + 4 - 4a + 64 = 25 + a^2 + 1 + 2a\\[1em] ⇒ a^2 - 4a + 68 = a^2 + 2a + 26\\[1em] ⇒ - 4a + 68 = 2a + 26\\[1em] ⇒ 68 - 26 = 2a + 4a\\[1em] ⇒ 6a = 42\\[1em] ⇒ a = \dfrac{42}{6}\\[1em] ⇒ a = 7

Hence, the value of a = 7.

Question 9

What point on the x-axis is equidistant from the points (7, 6) and (-3, 4) ?

Answer

Let the required point on the x-axis be (x, 0).

Given (x, 0) is equidistant from (7, 6) and (-3, 4).

i.e. distance between (x, 0) and (7, 6) = distance between (x, 0) and (-3, 4)

(7−x)2+(6−0)2=((−3)−x)2+(4−0)2⇒(7−x)2+(6)2=(−3−x)2+(4)2⇒(7−x)2+(6)2=(−3−x)2+(4)2⇒49+x2−14x+36=9+x2+6x+16⇒x2−14x+85=x2+6x+25⇒−14x+85=6x+25⇒−14x−6x=25−85⇒−20x=−60⇒x=6020⇒x=3\sqrt{(7 - x)^2 + (6 - 0)^2} = \sqrt{((-3) - x)^2 + (4 - 0)^2}\\[1em] ⇒ \sqrt{(7 - x)^2 + (6)^2} = \sqrt{(-3 - x)^2 + (4)^2}\\[1em] ⇒ (7 - x)^2 + (6)^2 = (-3 - x)^2 + (4)^2\\[1em] ⇒ 49 + x^2 - 14x + 36 = 9 + x^2 + 6x + 16\\[1em] ⇒ x^2 - 14x + 85 = x^2 + 6x + 25\\[1em] ⇒ - 14x + 85 = 6x + 25\\[1em] ⇒ - 14x - 6x = 25 - 85\\[1em] ⇒ - 20x = -60\\[1em] ⇒ x = \dfrac{60}{20}\\[1em] ⇒ x = 3

Hence, the point on the x-axis which is equidistant from the points (7, 6) and (-3, 4) is (3, 0).

Question 10

Find a point on the y-axis which is equidistant from the points (5, 2) and (-4, 3).

Answer

Let the required point on the y-axis be (0, y).

Given (0, y) is equidistant from (5, 2) and (-4, 3).

i.e. distance between (0, y) and (5, 2) = distance between (0, y) and (-4, 3)

(5−0)2+(2−y)2=((−4)−0)2+(3−y)2⇒(5)2+(2−y)2=(−4)2+(3−y)2⇒(5)2+(2−y)2=(−4)2+(3−y)2⇒25+4+y2−4y=16+9+y2−6y⇒29+y2−4y=25+y2−6y⇒29−4y=25−6y⇒−4y+6y=25−29⇒2y=−4⇒y=−42⇒y=−2\sqrt{(5 - 0)^2 + (2 - y)^2} = \sqrt{((-4) - 0)^2 + (3 - y)^2}\\[1em] ⇒ \sqrt{(5)^2 + (2 - y)^2} = \sqrt{(-4)^2 + (3 - y)^2}\\[1em] ⇒ (5)^2 + (2 - y)^2 = (-4)^2 + (3 - y)^2\\[1em] ⇒ 25 + 4 + y^2 - 4y = 16 + 9 + y^2 - 6y\\[1em] ⇒ 29 + y^2 - 4y = 25 + y^2 - 6y\\[1em] ⇒ 29 - 4y = 25 - 6y\\[1em] ⇒ - 4y + 6y = 25 - 29\\[1em] ⇒ 2y = - 4\\[1em] ⇒ y = - \dfrac{4}{2}\\[1em] ⇒ y = - 2

Hence, the point on the y-axis which is equidistant from the points (5, 2) and (-4, 3) is (0, -2).

Question 11

A point P lies on the x-axis and another point Q lies on the y-axis.

(i) Write the ordinate of point P.

(ii) Write the abscissa of point Q.

(iii) If the abscissa of point P is -12 and the ordinate of point Q is -16; calculate the length of line segment PQ.

Answer

(i) The point P lies on the x-axis, so its co-ordinates are (x, 0).

The ordinate of point P is 0.

(ii) The point Q lies on the y-axis, so its co-ordinates are (0, y).

The abscissa of point Q is 0.

(iii) If the abscissa of point P is -12, then P = (-12, 0).

And if the ordinate of point Q is -16. So, then Q = (0, -16).

The length of the line segment PQ

=(x2−x1)2+(y2−y1)2=(0−(−12))2+((−16)−0)2=(−12)2+(−16)2=144+256=400=20= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] = \sqrt{(0 - (-12))^2 + ((-16) - 0)^2}\\[1em] = \sqrt{(-12)^2 + (-16)^2}\\[1em] = \sqrt{144 + 256}\\[1em] = \sqrt{400}\\[1em] = 20

Hence, the length of PQ = 20.

Question 12

Show that the points P (0, 5), Q (5, 10) and R (6, 3) are the vertices of an isosceles triangle.

Answer

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

The length of PQ

=(5−0)2+(10−5)2=52+52=25+25=50=52= \sqrt{(5 - 0)^2 + (10 - 5)^2}\\[1em] = \sqrt{5^2 + 5^2}\\[1em] = \sqrt{25 + 25}\\[1em] = \sqrt{50}\\[1em] = 5\sqrt{2}\\[1em]

The length of QR

=(6−5)2+(3−10)2=12+(−7)2=1+49=50=52= \sqrt{(6 - 5)^2 + (3 - 10)^2}\\[1em] = \sqrt{1^2 + (-7)^2}\\[1em] = \sqrt{1 + 49}\\[1em] = \sqrt{50}\\[1em] = 5\sqrt{2}\\[1em]

The length of RP

=(6−0)2+(3−5)2=62+(−2)2=36+4=40=210= \sqrt{(6 - 0)^2 + (3 - 5)^2}\\[1em] = \sqrt{6^2 + (-2)^2}\\[1em] = \sqrt{36 + 4}\\[1em] = \sqrt{40}\\[1em] = 2\sqrt{10}\\[1em]

PQ = QR ⇒ the triangle is isosceles triangle

Hence, the triangle PQR is an isosceles triangle.

Question 13

Prove that the points P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1) are the vertices of a rectangle PQRS.

Answer

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

The length of PQ

=(6−0)2+(2−(−4))2=62+62=36+36=72=62= \sqrt{(6 - 0)^2 + (2 - (-4))^2}\\[1em] = \sqrt{6^2 + 6^2}\\[1em] = \sqrt{36 + 36}\\[1em] = \sqrt{72}\\[1em] = 6\sqrt{2}\\[1em]

The length of RS

=(−3−3)2+(−1−5)2=−62+(−6)2=36+36=72=62= \sqrt{(-3 - 3)^2 + (-1 - 5)^2}\\[1em] = \sqrt{-6^2 + (-6)^2}\\[1em] = \sqrt{36 + 36}\\[1em] = \sqrt{72}\\[1em] = 6\sqrt{2}\\[1em]

The length of QR

=(3−6)2+(5−2)2=(−3)2+(−3)2=9+9=18=32= \sqrt{(3 - 6)^2 + (5 - 2)^2}\\[1em] = \sqrt{(-3)^2 + (-3)^2}\\[1em] = \sqrt{9 + 9}\\[1em] = \sqrt{18}\\[1em] = 3\sqrt{2}\\[1em]

The length of SP

=(−3−0)2+(−1−(−4))2=−32+(−3)2=9+9=18=32= \sqrt{(-3 - 0)^2 + (-1 - (-4))^2}\\[1em] = \sqrt{-3^2 + (-3)^2}\\[1em] = \sqrt{9 + 9}\\[1em] = \sqrt{18}\\[1em] = 3\sqrt{2}\\[1em]

PQ = RS
QR = SP

The length of diagonal QS =

=(−3−6)2+(−1−2)2=(−9)2+(−3)2=81+9=90=310= \sqrt{(-3 - 6)^2 + (-1 - 2)^2}\\[1em] = \sqrt{(-9)^2 + (-3)^2}\\[1em] = \sqrt{81 + 9}\\[1em] = \sqrt{90}\\[1em] = 3\sqrt{10}\\[1em]

The length of diagonal PR =

=(3−0)2+(5−(−4))2=32+92=9+81=90=310= \sqrt{(3 - 0)^2 + (5 - (-4))^2}\\[1em] = \sqrt{3^2 + 9^2}\\[1em] = \sqrt{9 + 81}\\[1em] = \sqrt{90}\\[1em] = 3\sqrt{10}\\[1em]

So, QS = PR

Since opposite sides are equal, and the diagonals are equal, we can conclude that the quadrilateral PQRS is a rectangle.

Hence, the points P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1) are the vertices of a rectangle PQRS.

Question 14

Prove that the points A (1, -3), B (-3, 0) and C (4, 1) are the vertices of an isosceles right-angled triangle. Find the area of the triangle.

Answer

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

The length of AB

=(−3−1)2+(0−(−3))2=(−4)2+32=16+9=25=5= \sqrt{(-3 - 1)^2 + (0 - (-3))^2}\\[1em] = \sqrt{(-4)^2 + 3^2}\\[1em] = \sqrt{16 + 9}\\[1em] = \sqrt{25}\\[1em] = 5

The length of BC

=(4−(−3))2+(1−0)2=(−7)2+12=49+1=50=52= \sqrt{(4 - (-3))^2 + (1 - 0)^2}\\[1em] = \sqrt{(-7)^2 + 1^2}\\[1em] = \sqrt{49 + 1}\\[1em] = \sqrt{50}\\[1em] = 5\sqrt{2}\\[1em]

The length of CA

=(4−1)2+(1−(−3))2=32+42=9+16=25=5= \sqrt{(4 - 1)^2 + (1 - (-3))^2}\\[1em] = \sqrt{3^2 + 4^2}\\[1em] = \sqrt{9 + 16}\\[1em] = \sqrt{25}\\[1em] = 5

If ABC is an right angled triangle,

AB2 + CA2 = (25)(2\sqrt5)2 = 52 + 52 = 25 + 25 = 50 ⇒ BC2

BC2 = AB2 + CA2 ⇒ the triangle is right angled triangle.

and,

AB = CA ⇒ the triangle is isosceles triangle.

Base of triangle = Height of the triangle = 5 units.

Area of triangle ABC = 12\dfrac{1}{2} x base x height

=12×5×5=252=12.5 sq. units= \dfrac{1}{2} \times 5 \times 5\\[1em] = \dfrac{25}{2}\\[1em] = 12.5 \text{ sq. units}

Hence, the triangle ABC is an isosceles right-angled triangle and area of the triangle = 12.5 sq. units.

Question 15

Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square ABCD.

Answer

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

The length of AB

=(1−5)2+(5−6)2=(−4)2+(−1)2=16+1=17= \sqrt{(1 - 5)^2 + (5 - 6)^2}\\[1em] = \sqrt{(-4)^2 + (-1)^2}\\[1em] = \sqrt{16 + 1}\\[1em] = \sqrt{17}

The length of BC

=(2−1)2+(1−5)2=12+(−4)2=1+16=17= \sqrt{(2 - 1)^2 + (1 - 5)^2}\\[1em] = \sqrt{1^2 + (-4)^2}\\[1em] = \sqrt{1 + 16}\\[1em] = \sqrt{17}

The length of CD

=(6−2)2+(2−1)2=42+12=16+1=17= \sqrt{(6 - 2)^2 + (2 - 1)^2}\\[1em] = \sqrt{4^2 + 1^2}\\[1em] = \sqrt{16 + 1}\\[1em] = \sqrt{17}

The length of DA

=(6−5)2+(2−6)2=12+42=1+16=17= \sqrt{(6 - 5)^2 + (2 - 6)^2}\\[1em] = \sqrt{1^2 + 4^2}\\[1em] = \sqrt{1 + 16}\\[1em] = \sqrt{17}

AB = BC = CD = DA = 17\sqrt{17}

The length of diagonal AC =

=(2−5)2+(1−6)2=(−3)2+(−5)2=9+25=34= \sqrt{(2 - 5)^2 + (1 - 6)^2}\\[1em] = \sqrt{(-3)^2 + (-5)^2}\\[1em] = \sqrt{9 + 25}\\[1em] = \sqrt{34}

The length of diagonal BD =

=(6−1)2+(2−5)2=(−5)2+(−3)2=25+9=34= \sqrt{(6 - 1)^2 + (2 - 5)^2}\\[1em] = \sqrt{(-5)^2 + (-3)^2}\\[1em] = \sqrt{25 + 9}\\[1em] = \sqrt{34}

So, AC = BD

Since all sides and diagonals are equal, the points form a square.

Hence, the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square ABCD.

Test Yourself

Question 1(a)

The distance of point (-8, 6) from x-axis is:

  1. 8

  2. 6

  3. 10

  4. none of these

Answer

Plot the point P(-8, 6) on the graph paper.

Draw a perpendicular line from the point P to the x-axis.

From graph, it is clear that the distance between the point (-8, 6) and the x-axis is 6 units.

The distance of point (-8, 6) from x-axis is: Distance Formula, Concise Mathematics Solutions ICSE Class 9.

Hence, option 2 is the correct option.

Question 1(b)

The distance of point (-4, -3) from the origin is:

  1. -10 unit

  2. 10 unit

  3. 42−32\sqrt{4^2 - 3^2}

  4. none of these

Answer

Origin (O) = (0, 0) and P = (-4, -3).

By distance formula,

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

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

Hence, option 4 is the correct option.

Question 1(c)

The co-ordinates of point P are:

The co-ordinates of point P are: Distance Formula, Concise Mathematics Solutions ICSE Class 9.

  1. (0, 5)

  2. (5, 0)

  3. (4, 3)

  4. (3, 4)

Answer

The co-ordinates of point P are: Distance Formula, Concise Mathematics Solutions ICSE Class 9.

Since, OPQ is right angled triangle. Using pythagoras theorem,

⇒ Hypotenuse2 = Base2 + Height2

⇒ OP2 = OQ2 + QP2

⇒ 52 = 32 + QP2

⇒ 25 = 9 + QP2

⇒ QP2 = 25 - 9

⇒ QP2 = 16

⇒ QP = 16\sqrt{16}

⇒ QP = 4 units

Since, OQ = 3 units and QP = 4 units.

The co-ordinates of point P are (3, 4).

Hence, option 4 is the correct option.

Question 1(d)

AB (= 10 unit) is diameter of a circle with center at point P = (x, 0) and point B = (0, y). The relation between x and y is:

AB (= 10 unit) is diameter of a circle with center at point P = (x, 0) and point B = (0, y). The relation between x and y is: Distance Formula, Concise Mathematics Solutions ICSE Class 9.

  1. x + y = 10

  2. x + y = 25

  3. x2 + y2 = 5

  4. x2 + y2 = 25

Answer

Given, AB is diameter of a circle with center at point P = (x, 0) and point B = (0, y).

By distance formula,

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

From figure,

AB (Diameter) = 2 x PB (Radius)

⇒10=2×(x−0)2+(0−y)2⇒102=x2+(−y)2⇒5=x2+y2⇒52=x2+y2⇒x2+y2=25.\Rightarrow 10 = 2 \times \sqrt{(x - 0)^2 + (0 - y)^2}\\[1em] \Rightarrow \dfrac{10}{2} = \sqrt{x^2 + (-y)^2}\\[1em] \Rightarrow 5 = \sqrt{x^2 + y^2}\\[1em] \Rightarrow 5^2 = x^2 + y^2\\[1em] \Rightarrow x^2 + y^2 = 25.

Hence, option 4 is the correct option.

Question 1(e)

Statement 1: For the point P, x = -4 and y = 3, the distance of P from origin is 3 + 4 = 7.

Statement 2: P = (-4, 3) and its distance from origin = (−4)2+(3)2\sqrt{(-4)^2 + (3)^2}.

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Answer

For the point P, x = -4 and y = 3, the distance of P from origin is 3 + 4 = 7. Distance Formula, Concise Mathematics Solutions ICSE Class 9.

Given, point P, x = -4 and y = 3.

Using distance formula,

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

Distance of P(-4, 3) from origin (0, 0)

=(0−(−4))2+(0−3)2=42+(−3)2=16+9=25=5 units.= \sqrt{(0 - (-4))^2 + (0 - 3)^2}\\[1em] = \sqrt{4^2 + (-3)^2}\\[1em] = \sqrt{16 + 9}\\[1em] = \sqrt{25}\\[1em] = 5 \text{ units}.

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Question 1(f)

Statement 1: The point P(x, y) is at a distance of 6 unit from origin, then P lies in the first quadrant.

Statement 2: Point P can lie in any quadrant.

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Answer

Given, the point P(x, y) is at a distance of 6 unit from origin.

By distance formula,

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

⇒6=(x−0)2+(y−0)2⇒6=(x)2+(y)2⇒62=x2+y2⇒x2+y2=36.\Rightarrow 6 = \sqrt{(x - 0)^2 + (y - 0)^2}\\[1em] \Rightarrow 6 = \sqrt{(x)^2 + (y)^2}\\[1em] \Rightarrow 6^2 = x^2 + y^2\\[1em] \Rightarrow x^2 + y^2 = 36.

The above equation defines a circle centered at the origin with radius 6. That circle covers all four quadrants, so P can lie anywhere on that circle, not necessarily in the first quadrant.

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Question 1(g)

Assertion (A): If A = (2x, y), B(x, 2y) and AB = 5 unit, then x + y = 5.

Reason (R): (x−2x)2+(2y−y)2\sqrt{(x - 2x)^2 + (2y - y)^2} = 5

⇒ x2 + y2 = 25

  1. A is true, but R is false.

  2. A is false, but R is true.

  3. Both A and R are true, and R is the correct reason for A.

  4. Both A and R are true, and R is the incorrect reason for A.

Answer

Given, A = (2x, y), B = (x, 2y).

By distance formula,

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

Since, AB = 5 unit.

Substituting the values, we get :

⇒5=(x−2x)2+(2y−y)2⇒5=(−x)2+y2⇒52=x2+y2⇒x2+y2=25.\Rightarrow 5 = \sqrt{(x - 2x)^2 + (2y - y)^2}\\[1em] \Rightarrow 5 = \sqrt{(-x)^2 + y^2}\\[1em] \Rightarrow 5^2 = x^2 + y^2\\[1em] \Rightarrow x^2 + y^2 = 25.

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Question 1(h)

Assertion (A): The distance between the points A(x, 2x) and B(x, 0) is 4 unit, the point B is (2, 0).

Reason (R): (x−x)2+(2x−0)2\sqrt{(x - x)^2 + (2x - 0)^2} = 4

⇒ x2 = 4 and x = ± 2

  1. A is true, but R is false.

  2. A is false, but R is true.

  3. Both A and R are true, and R is the correct reason for A.

  4. Both A and R are true, and R is the incorrect reason for A.

Answer

Given, the distance between the points A(x, 2x) and B(x, 0) is 4 unit.

By distance formula,

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

Substituting the values, we get :

⇒4=(x−x)2+(2x−0)2⇒4=02+(2x)2⇒4=4x2⇒42=4x2⇒16=4x2⇒x2=164⇒x2=4⇒x=4⇒x=±2.\Rightarrow 4 = \sqrt{(x - x)^2 + (2x - 0)^2}\\[1em] \Rightarrow 4 = \sqrt{0^2 + (2x)^2}\\[1em] \Rightarrow 4 = \sqrt{4x^2}\\[1em] \Rightarrow 4^2 = 4x^2\\[1em] \Rightarrow 16 = 4x^2\\[1em] \Rightarrow x^2 = \dfrac{16}{4}\\[1em] \Rightarrow x^2 = 4\\[1em] \Rightarrow x = \sqrt{4}\\[1em] \Rightarrow x = \pm 2.

The coordinates of B = (x, 0) = (2, 0) or (-2, 0)

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Question 2

Find the points on the y-axis which are at a distance of 252{\sqrt5} units from the point (-4, 7).

Answer

Let the point on y-axis be (0, y).

Let (0, y) = (x1, y1) and (-4, 7) = (x2, y2)

⇒ Distance between the given points =

(x2−x1)2+(y2−y1)2⇒25=(−4−0)2+(7−y)2⇒25=(−4)2+(7−y)2⇒(25)2=(−4)2+(7−y)2⇒20=16+49+y2−14y⇒20=65+y2−14y⇒65+y2−14y−20=0⇒y2−14y+45=0⇒y2−(9y+5y)+45=0⇒y2−9y−5y+45=0⇒(y2−9y)−(5y−45)=0⇒y(y−9)−5(y−9)=0⇒(y−9)(y−5)=0⇒y=9 or 5\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] ⇒ 2 \sqrt5 = \sqrt{(-4 - 0)^2 + (7 - y)^2}\\[1em] ⇒ 2 \sqrt5 = \sqrt{(-4)^2 + (7 - y)^2}\\[1em] ⇒ (2 \sqrt5)^2 = (-4)^2 + (7 - y)^2\\[1em] ⇒ 20 = 16 + 49 + y^2 - 14y\\[1em] ⇒ 20 = 65 + y^2 - 14y\\[1em] ⇒ 65 + y^2 - 14y - 20 = 0\\[1em] ⇒ y^2 - 14y + 45 = 0\\[1em] ⇒ y^2 - (9y + 5y) + 45 = 0\\[1em] ⇒ y^2 - 9y - 5y + 45 = 0\\[1em] ⇒ (y^2 - 9y) - (5y - 45) = 0\\[1em] ⇒ y(y - 9) - 5(y - 9) = 0\\[1em] ⇒ (y - 9)(y - 5) = 0\\[1em] ⇒ y = 9 \text{ or } 5

Hence, the the points on the y-axis are (0, 9) or (0, 5).

Question 3

Find the value of k, if the points (5, k) and (k, 7) are equidistant from point (2, 4).

Answer

Given (2, 4) is equidistant from (5, k) and (k, 7).

i.e. distance between (2, 4) and (5, k) = distance between (2, 4) and (k, 7)

(2−5)2+(4−k)2=(2−k)2+(4−7)2⇒(2−5)2+(4−k)2=(2−k)2+(4−7)2⇒(−3)2+(4−k)2=(2−k)2+(−3)2⇒9+16+k2−8k=4+k2−4k+9⇒25+k2−8k=13+k2−4k⇒13+k2−4k−25−k2+8k=0⇒−12+4k=0⇒4k=12⇒k=124⇒k=3\sqrt{(2 - 5)^2 + (4 - k)^2} = \sqrt{(2 - k)^2 + (4 - 7)^2}\\[1em] ⇒ (2 - 5)^2 + (4 - k)^2 = (2 - k)^2 + (4 - 7)^2\\[1em] ⇒ (- 3)^2 + (4 - k)^2 = (2 - k)^2 + (- 3)^2\\[1em] ⇒ 9 + 16 + k^2 - 8k = 4 + k^2 - 4k + 9\\[1em] ⇒ 25 + k^2 - 8k = 13 + k^2 - 4k\\[1em] ⇒ 13 + k^2 - 4k - 25 - k^2 + 8k = 0\\[1em] ⇒ - 12 + 4k = 0\\[1em] ⇒ 4k = 12\\[1em] ⇒ k = \dfrac{12}{4}\\[1em] ⇒ k = 3

Hence, the value of k = 3.

Question 4

The centre of a circle is (2a, a - 7). Find the value (values) of a, if the circle passes through the point (11, -9) and has diameter 10210{\sqrt2} units.

Answer

Radius of circle = Diameter2\dfrac{\text{Diameter}}{2}

= 1022\dfrac{10 \sqrt2}{2}

= 525 \sqrt2

⇒ Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Radius of the circle = The distance between the center (2a, a - 7) and the point (11, -9), which lies on the circle

52=(2a−11)2+((a−7)−(−9))2⇒(52)2=(2a−11)2+(a−7+9)2⇒(52)2=(2a−11)2+(a+2)2⇒50=4a2+121−44a+a2+4+4a⇒50=5a2+125−40a⇒5a2+125−40a−50=0⇒5a2−40a+75=0⇒a2−8a+15=0⇒a2−(3a+5a)+15=0⇒a2−3a−5a+15=0⇒a(a−3)−5(a−3)=0⇒(a−3)(a−5)=0⇒a=3 or 55 \sqrt2 = \sqrt{(2a - 11)^2 + ((a - 7) - (-9))^2}\\[1em] ⇒ (5 \sqrt2)^2 = (2a - 11)^2 + (a - 7 + 9)^2\\[1em] ⇒ (5 \sqrt2)^2 = (2a - 11)^2 + (a + 2)^2\\[1em] ⇒ 50 = 4a^2 + 121 - 44a + a^2 + 4 + 4a\\[1em] ⇒ 50 = 5a^2 + 125 - 40a\\[1em] ⇒ 5a^2 + 125 - 40a - 50 = 0\\[1em] ⇒ 5a^2 - 40a + 75 = 0\\[1em] ⇒ a^2 - 8a + 15 = 0\\[1em] ⇒ a^2 - (3a + 5a) + 15 = 0\\[1em] ⇒ a^2 - 3a - 5a + 15 = 0\\[1em] ⇒ a(a - 3) - 5(a - 3) = 0\\[1em] ⇒ (a - 3)(a - 5) = 0\\[1em] ⇒ a = 3 \text{ or } 5

Hence, the value of a = 3 or a = 5.

Question 5

Show that (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.

Answer

The points A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4) are the vertices of a quadrilateral. We need to show that this quadrilateral is a rhombus.

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Distance between points A(-3, 2) and B(-5, -5):

=((−5)−(−3))2+((−5)−2)2=(−2)2+(−7)2=4+49=53= \sqrt{((-5) - (-3))^2 + ((-5) - 2)^2}\\[1em] = \sqrt{(-2)^2 + (-7)^2}\\[1em] = \sqrt{4 + 49}\\[1em] = \sqrt{53}

Distance between points B(-5, -5) and C(2, -3):

=(2−(−5))2+((−3)−(−5))2=72+22=49+4=53= \sqrt{(2 - (-5))^2 + ((-3) - (-5))^2}\\[1em] = \sqrt{7^2 + 2^2}\\[1em] = \sqrt{49 + 4}\\[1em] = \sqrt{53}

Distance between points C(2, -3) and D(4, 4):

=(4−2)2+(4−(−3))2=22+72=4+49=53= \sqrt{(4 - 2)^2 + (4 - (-3))^2}\\[1em] = \sqrt{2^2 + 7^2}\\[1em] = \sqrt{4 + 49}\\[1em] = \sqrt{53}

Distance between points D(4, 4) and A(-3, 2):

=(4−2)2+(4−(−3))2=22+72=4+49=53= \sqrt{(4 - 2)^2 + (4 - (-3))^2}\\[1em] = \sqrt{2^2 + 7^2}\\[1em] = \sqrt{4 + 49}\\[1em] = \sqrt{53}

AB = BC = CD = DA = 53\sqrt{53}

Since all sides are equal, the quadrilateral is a rhombus.

Hence, the points (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.

Question 6

Points A (-3, -2), B (-6, a), C (-3, -4) and D(0, -1) are the vertices of quadrilateral ABCD; find a if 'a' is negative and AB = CD.

Answer

Given:

Points A(-3, -2), B(-6, a), C(-3, -4) and D(0, -1) are the vertices of quadrilateral ABCD and AB = CD.

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Distance between points A(-3, -2) and B(-6, a):

=((−6)−(−3))2+(a−(−2))2=(−3)2+(a−(−2))2=9+a2+4+4a=13+a2+4a= \sqrt{((-6) - (-3))^2 + (a - (-2))^2}\\[1em] = \sqrt{(-3)^2 + (a - (-2))^2}\\[1em] = \sqrt{9 + a^2 + 4 + 4a}\\[1em] = \sqrt{13 + a^2 + 4a}

Distance between points C (-3, -4) and D(0, -1):

=(0−(−3))2+((−1)−(−4))2=32+32=9+9=18= \sqrt{(0 - (-3))^2 + ((-1) - (-4))^2}\\[1em] = \sqrt{3^2 + 3^2}\\[1em] = \sqrt{9 + 9}\\[1em] = \sqrt{18}

Since AB = CD,

⇒13+a2+4a=18⇒13+a2+4a=18⇒13+a2+4a−18=0⇒a2+4a−5=0⇒a2+5a−1a−5=0⇒(a2+5a)−(1a+5)=0⇒a(a+5)−1(a+5)=0⇒(a+5)(a−1)=0⇒a=−5 or 1⇒ \sqrt{13 + a^2 + 4a} = \sqrt{18}\\[1em] ⇒ 13 + a^2 + 4a = 18\\[1em] ⇒ 13 + a^2 + 4a - 18 = 0\\[1em] ⇒ a^2 + 4a - 5 = 0\\[1em] ⇒ a^2 + 5a - 1a - 5 = 0\\[1em] ⇒ (a^2 + 5a) - (1a + 5) = 0\\[1em] ⇒ a(a + 5) - 1(a + 5) = 0\\[1em] ⇒ (a + 5)(a - 1) = 0\\[1em] ⇒ a = -5 \text{ or } 1

Since it is given that a is negative, we select a = -5.

Hence, the value of a = -5.

Question 7

The vertices of a triangle are (5, 1), (11, 1) and (11, 9). Find the co-ordinates of the circumcentre of the triangle.

Answer

Let the circumcentre of the triangle be P(x, y).

The circumcentre is equidistant from all three vertices of the triangle.

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

The distance from P to A is equal to the distance from P to B, so:

⇒(x−5)2+(y−1)2=(x−11)2+(y−1)2⇒(x−5)2+(y−1)2=(x−11)2+(y−1)2⇒x2+25−10x+y2+1−2y=x2+121−22x+y2+1−2y⇒x2+26−10x+y2−2y=x2+122−22x+y2−2y⇒26−10x=122−22x⇒22x−10x=122−26⇒12x=96⇒x=9612⇒x=8⇒ \sqrt{(x - 5)^2 + (y - 1)^2} = \sqrt{(x - 11)^2 + (y - 1)^2}\\[1em] ⇒ (x - 5)^2 + (y - 1)^2 = (x - 11)^2 + (y - 1)^2\\[1em] ⇒ x^2 + 25 - 10x + y^2 + 1 - 2y = x^2 + 121 - 22x + y^2 + 1 - 2y\\[1em] ⇒ x^2 + 26 - 10x + y^2 - 2y = x^2 + 122 - 22x + y^2 - 2y\\[1em] ⇒ 26 - 10x = 122 - 22x\\[1em] ⇒ 22x - 10x = 122 - 26\\[1em] ⇒ 12x = 96\\[1em] ⇒ x = \dfrac{96}{12}\\[1em] ⇒ x = 8

The distance from P to A is equal to the distance from P to C, so:

⇒(x−5)2+(y−1)2=(x−11)2+(y−9)2⇒(x−5)2+(y−1)2=(x−11)2+(y−9)2⇒x2+25−10x+y2+1−2y=x2+121−22x+y2+81−18y⇒x2+26−10x+y2−2y=x2+202−22x+y2−18y⇒26−10x−2y=202−22x−18y⇒26−10x−2y−202+22x+18y=0⇒12x+16y=176⇒3x+4y=44⇒ \sqrt{(x - 5)^2 + (y - 1)^2} = \sqrt{(x - 11)^2 + (y - 9)^2}\\[1em] ⇒ (x - 5)^2 + (y - 1)^2 = (x - 11)^2 + (y - 9)^2\\[1em] ⇒ x^2 + 25 - 10x + y^2 + 1 - 2y = x^2 + 121 - 22x + y^2 + 81 - 18y\\[1em] ⇒ x^2 + 26 - 10x + y^2 - 2y = x^2 + 202 - 22x + y^2 - 18y\\[1em] ⇒ 26 - 10x - 2y = 202 - 22x - 18y\\[1em] ⇒ 26 - 10x - 2y - 202 + 22x + 18y = 0\\[1em] ⇒ 12x + 16y = 176\\[1em] ⇒ 3x + 4y = 44\\[1em]

Putting the value of x = 8 in the above equation,

⇒3×8+4y=44⇒24+4y=44⇒4y=44−24⇒4y=20⇒y=204⇒y=5⇒ 3 \times 8 + 4y = 44\\[1em] ⇒ 24 + 4y = 44\\[1em] ⇒ 4y = 44 - 24\\[1em] ⇒ 4y = 20\\[1em] ⇒ y = \dfrac{20}{4}\\[1em] ⇒ y = 5

Hence, the co-ordinates of the circumcentre of the triangle is (8, 5).

Question 8

Given A = (3, 1) and B = (0, y - 1). Find y if AB = 5.

Answer

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Distance between A(3, 1) and B(0, y - 1):

⇒(0−3)2+((y−1)−1)2=5⇒(−3)2+(y−1−1)2=52⇒9+(y−2)2=25⇒9+y2+4−4y=25⇒y2−4y+13=25⇒y2−4y−25+13=0⇒y2−4y−12=0⇒y2−6y+2y−12=0⇒(y2−6y)+(2y−12)=0⇒y(y−6)+2(y−6)=0⇒(y−6)(y+2)=0⇒y=6 or −2⇒ \sqrt{(0 - 3)^2 + ((y - 1) - 1)^2} = 5\\[1em] ⇒ (-3)^2 + (y - 1 - 1)^2 = 5^2\\[1em] ⇒ 9 + (y - 2)^2 = 25\\[1em] ⇒ 9 + y^2 + 4 - 4y = 25\\[1em] ⇒ y^2 - 4y + 13 = 25\\[1em] ⇒ y^2 - 4y - 25 + 13 = 0\\[1em] ⇒ y^2 - 4y - 12 = 0\\[1em] ⇒ y^2 - 6y + 2y - 12 = 0\\[1em] ⇒ (y^2 - 6y) + (2y - 12) = 0\\[1em] ⇒ y(y - 6) + 2(y - 6) = 0\\[1em] ⇒ (y - 6)(y + 2) = 0\\[1em] ⇒ y = 6 \text{ or } -2

Hence, the values of y are 6 and -2.

Question 9

Given A = (x + 2, -2) and B = (11, 6). Find x if AB = 17.

Answer

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Distance between A(x + 2, -2) and B(11, 6):

⇒(11−(x+2))2+(6−(−2))2=17⇒(11−x−2)2+(6+2)2=172⇒(9−x)2+82=289⇒81+x2−18x+64=289⇒x2−18x+145=289⇒x2−18x+145−289=0⇒x2−18x−144=0⇒x2−24x+6x−144=0⇒(x2−24x)+(6x−144)=0⇒x(x−24)+6(x−24)=0⇒(x−24)(x+6)=0⇒x=24 and −6⇒ \sqrt{(11 - (x + 2))^2 + (6 - (-2))^2} = 17\\[1em] ⇒ (11 - x - 2)^2 + (6 + 2)^2 = 17^2\\[1em] ⇒ (9 - x)^2 + 8^2 = 289\\[1em] ⇒ 81 + x^2 - 18x + 64 = 289\\[1em] ⇒ x^2 - 18x + 145 = 289\\[1em] ⇒ x^2 - 18x + 145 - 289 = 0\\[1em] ⇒ x^2 - 18x - 144 = 0\\[1em] ⇒ x^2 - 24x + 6x - 144 = 0\\[1em] ⇒ (x^2 - 24x) + (6x - 144) = 0\\[1em] ⇒ x(x - 24) + 6(x - 24) = 0\\[1em] ⇒ (x - 24)(x + 6) = 0\\[1em] ⇒ x = 24 \text{ and } -6

Hence, the values of x are 24 and -6.

Question 10

The centre of a circle is (2x - 1, 3x + 1). Find x if the circle passes through (-3, -1) and the length of its diameter is 20 units.

Answer

The diameter of the circle is given as 20 units, so the radius is 10 units.

Distance between the centre A (2x - 1, 3x + 1) and point B (-3, -1) = Radius of circle AB = 10

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

⇒ AB2 = 100

Distance between B(-3, -1) and A(2x - 1, 3x + 1):

⇒((2x−1)−(−3))2+((3x+1)−(−1))2=100⇒(2x−1+3)2+(3x+1+1)2=100⇒(2x+2)2+(3x+2)2=100⇒4x2+4+8x+9x2+4+12x=100⇒13x2+8+20x=100⇒13x2+8+20x−100=0⇒13x2+20x−92=0⇒x=−20+400+478426 or −20−400+478426⇒x=−20+518426 or −20−518426⇒x=−20+7226 or −20−7226⇒x=5226 or −9226⇒x=2 or −4613⇒ ((2x - 1) - (-3))^2 + ((3x + 1) - (-1))^2 = 100\\[1em] ⇒ (2x - 1 + 3)^2 + (3x + 1 + 1)^2 = 100\\[1em] ⇒ (2x + 2)^2 + (3x + 2)^2 = 100\\[1em] ⇒ 4x^2 + 4 + 8x + 9x^2 + 4 + 12x = 100\\[1em] ⇒ 13x^2 + 8 + 20x = 100\\[1em] ⇒ 13x^2 + 8 + 20x - 100 = 0\\[1em] ⇒ 13x^2 + 20x - 92 = 0\\[1em] ⇒ x = \dfrac{-20 + \sqrt{400 + 4784}}{26} \text { or } \dfrac{-20 - \sqrt{400 + 4784}}{26}\\[1em] ⇒ x = \dfrac{-20 + \sqrt{5184}}{26} \text { or } \dfrac{-20 - \sqrt{5184}}{26}\\[1em] ⇒ x = \dfrac{-20 + 72}{26} \text { or } \dfrac{-20 - 72}{26}\\[1em] ⇒ x = \dfrac{52}{26} \text { or } \dfrac{-92}{26}\\[1em] ⇒ x = 2 \text { or } \dfrac{-46}{13}\\[1em]

Hence, the values of x are 2 and −4613\dfrac{-46}{13}.

Question 11

The length of line PQ is 10 units and the co-ordinates of P are (2, -3); calculate the co-ordinates of point Q, if its abscissa is 10.

Answer

Let the co-ordinates of point Q be (10, y).

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Distance between P(2, -3) and Q(10, y):

⇒(10−2)2+(y−(−3))2=10⇒(10−2)2+(y−(−3))2=100⇒82+(y+3)2=100⇒64+y2+9+6y=100⇒y2+6y+73=100⇒y2+6y+73−100=0⇒y2+6y−27=0⇒y2+9y−3y−27=0⇒(y2+9y)−(3y+27)=0⇒y(y+9)−3(y+9)=0⇒(y+9)(y−3)=0⇒y=−9 or 3⇒ \sqrt{(10 - 2)^2 + (y - (-3))^2} = 10\\[1em] ⇒ (10 - 2)^2 + (y - (-3))^2 = 100\\[1em] ⇒ 8^2 + (y + 3)^2 = 100\\[1em] ⇒ 64 + y^2 + 9 + 6y = 100\\[1em] ⇒ y^2 + 6y + 73 = 100\\[1em] ⇒ y^2 + 6y + 73 - 100 = 0\\[1em] ⇒ y^2 + 6y - 27 = 0\\[1em] ⇒ y^2 + 9y - 3y - 27 = 0\\[1em] ⇒ (y^2 + 9y) - (3y + 27) = 0\\[1em] ⇒ y(y + 9) - 3(y + 9) = 0\\[1em] ⇒ (y + 9)(y - 3) = 0\\[1em] ⇒ y = -9 \text{ or } 3

Hence, the required co-ordinates of the point Q are (10, -9) and (10, 3).

Question 12

Point P (2, -7) is the centre of a circle with radius 13 units, PT is perpendicular to chord AB and T = (-2, -4); Calculate the length of :

(i) AT

(ii) AB.

Point P (2, -7) is the centre of a circle with radius 13 units, PT is perpendicular to chord AB and T = (-2, -4); Calculate the length of : Distance Formula, Concise Mathematics Solutions ICSE Class 9.

Answer

(i) Given:

Radius = PA = PB = 13 units

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Distance between P(2, -7) and T(-2, -4):

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

Using Pythagoras theorem in triangle PAT,

PA2 = PT2 + AT2

⇒ AT2 = PA2 - PT2

⇒ AT2 = 132 - 52

⇒ AT2 = 169 - 25

⇒ AT2 = 144

⇒ AT = 144\sqrt{144}

⇒ AT = 12 units

Hence, the value of AT = 12 units.

(ii) We know that the perpendicular from the center of a circle to a chord bisects the chord.

AB = 2AT

= 2 x 12 units

= 24 units

Hence, the length of AB = 24 units.

Question 13

Calculate the distance between the points P(2, 2) and Q(5, 4) correct to three significant figures.

Answer

∵ Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Distance between P(2, 2) and Q(5, 4):

=(5−2)2+(4−2)2=32+22=9+4=13=3.605 units= \sqrt{(5 - 2)^2 + (4 - 2)^2}\\[1em] = \sqrt{3^2 + 2^2}\\[1em] = \sqrt{9 + 4}\\[1em] = \sqrt{13}\\[1em] = 3.605 \text{ units}

Hence, the distance between the points P(2, 2) and Q(5, 4) is 3.605 units.

Question 14

Calculate the distance between A(7, 3) and B on the x-axis whose abscissa is 11.

Answer

We know that any point on x-axis has co-ordinates of the form (x, 0).

The abscissa of point B is 11.

So, the point B is (11, 0).

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Distance between A(7, 3) and B(11, 0):

=(11−7)2+(0−3)2=42+(−3)2=16+9=25=5 units= \sqrt{(11 - 7)^2 + (0 - 3)^2}\\[1em] = \sqrt{4^2 + (- 3)^2}\\[1em] = \sqrt{16 + 9}\\[1em] = \sqrt{25}\\[1em] = 5 \text{ units}

Hence, the distance between A(7, 3) and B(11, 0) is 5 units.

Question 15

Calculate the distance between A(5, -3) and B on the y-axis whose ordinate is 9.

Answer

We know that any point on the y-axis has co-ordinates of the form (0, y).

The ordinate of point B is 9.

So, the point B is (0, 9).

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Distance between A(5, -3) and B(0, 9):

=(0−5)2+(9−(−3))2=(−5)2+122=25+144=169=13= \sqrt{(0 - 5)^2 + (9 - (-3))^2}\\[1em] = \sqrt{(-5)^2 + 12^2}\\[1em] = \sqrt{25 + 144}\\[1em] = \sqrt{169}\\[1em] = 13

Hence, the distance between A(5, -3) and B(0, 9) is 13 units.

Question 16

Find the point on y-axis whose distances from the points A(6, 7) and B(4, -3) are in the ratio 1 : 2.

Answer

Let the point on y-axis be P(0, y).

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Distance between A(6, 7) and P(0, y):

=(0−6)2+(y−7)2=(−6)2+(y−7)2=36+y2+49−14y=y2−14y+85= \sqrt{(0 - 6)^2 + (y - 7)^2}\\[1em] = \sqrt{(- 6)^2 + (y - 7)^2}\\[1em] = \sqrt{36 + y^2 + 49 - 14y}\\[1em] = \sqrt{y^2 - 14y + 85}\\[1em]

Distance between B(4, -3) and P(0, y):

=(0−4)2+(y−(−3))2=(−4)2+(y+3)2=16+y2+9+6y=y2+6y+25= \sqrt{(0 - 4)^2 + (y - (-3))^2}\\[1em] = \sqrt{(- 4)^2 + (y + 3)^2}\\[1em] = \sqrt{16 + y^2 + 9 + 6y}\\[1em] = \sqrt{y^2 + 6y + 25}\\[1em]

It is given that the point on y-axis whose distances from the points A(6, 7) and B(4, -3) are in the ratio 1 : 2.

⇒PAPB=12⇒y2−14y+85y2+6y+25=12⇒y2−14y+85y2+6y+25=14⇒4(y2−14y+85)=y2+6y+25⇒4y2−56y+340=y2+6y+25⇒4y2−56y+340−y2−6y−25=0⇒3y2−62y+315=0⇒y=62+3844−37806 or 62−3844−37806⇒y=62+646 or 62−646⇒y=62+86 or 62−86⇒y=706 or 546⇒y=353 or 9⇒\dfrac{PA}{PB} = \dfrac{1}{2}\\[1em] ⇒\dfrac{\sqrt{y^2 - 14y + 85}}{\sqrt{y^2 + 6y + 25}} = \dfrac{1}{2}\\[1em] ⇒\dfrac{y^2 - 14y + 85}{y^2 + 6y + 25} = \dfrac{1}{4}\\[1em] ⇒4(y^2 - 14y + 85) = y^2 + 6y + 25\\[1em] ⇒ 4y^2 - 56y + 340 = y^2 + 6y + 25\\[1em] ⇒ 4y^2 - 56y + 340 - y^2 - 6y - 25 = 0\\[1em] ⇒ 3y^2 - 62y + 315 = 0\\[1em] ⇒ y = \dfrac{62 + \sqrt{3844 - 3780}}{6} \text { or } \dfrac{62 - \sqrt{3844 - 3780}}{6}\\[1em] ⇒ y = \dfrac{62 + \sqrt{64}}{6} \text { or } \dfrac{62 - \sqrt{64}}{6}\\[1em] ⇒ y = \dfrac{62 + 8}{6} \text { or } \dfrac{62 - 8}{6}\\[1em] ⇒ y = \dfrac{70}{6} \text { or } \dfrac{54}{6}\\[1em] ⇒ y = \dfrac{35}{3} \text { or } 9 \\[1em]

Hence, the required points on y-axis are (0, 9) and (0,353)\Big(0, \dfrac{35}{3}\Big).

Question 17

The distances of point P(x, y) from the points A(1, -3) and B(-2, 2) are in the ratio 2 : 3.

Show that : 5x2 + 5y2 - 34x + 70y + 58 = 0.

Answer

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Distance between A(1, -3) and P(x, y):

=(x−1)2+(y−(−3))2=(x−1)2+(y+3)2=x2+1−2x+y2+9+6y=x2−2x+y2+10+6y= \sqrt{(x - 1)^2 + (y - (-3))^2}\\[1em] = \sqrt{(x - 1)^2 + (y + 3)^2}\\[1em] = \sqrt{x^2 + 1 - 2x + y^2 + 9 + 6y}\\[1em] = \sqrt{x^2 - 2x + y^2 + 10 + 6y}\\[1em]

Distance between B(-2, 2) and P(x, y):

=(x−(−2))2+(y−2)2=(x+2)2+(y−2)2=x2+4+4x+y2+4−4y=x2+4x+y2+8−4y= \sqrt{(x - (-2))^2 + (y - 2)^2}\\[1em] = \sqrt{(x + 2)^2 + (y - 2)^2}\\[1em] = \sqrt{x^2 + 4 + 4x + y^2 + 4 - 4y}\\[1em] = \sqrt{x^2 + 4x + y^2 + 8 - 4y}\\[1em]

It is given that the distances of point P(x, y) from the points A(1, -3) and B(-2, 2) are in the ratio 2 : 3.

⇒PAPB=23⇒x2−2x+y2+10+6yx2+4x+y2+8−4y=23⇒x2−2x+y2+10+6yx2+4x+y2+8−4y=49⇒9(x2−2x+y2+10+6y)=4(x2+4x+y2+8−4y)⇒9x2−18x+9y2+90+54y=4x2+16x+4y2+32−16y⇒9x2−18x+9y2+90+54y−4x2−16x−4y2−32+16y=0⇒5x2+5y2−34x+70y+58=0⇒\dfrac{PA}{PB} = \dfrac{2}{3}\\[1em] ⇒\dfrac{\sqrt{x^2 - 2x + y^2 + 10 + 6y}}{\sqrt{x^2 + 4x + y^2 + 8 - 4y}} = \dfrac{2}{3}\\[1em] ⇒\dfrac{x^2 - 2x + y^2 + 10 + 6y}{x^2 + 4x + y^2 + 8 - 4y} = \dfrac{4}{9}\\[1em] ⇒9(x^2 - 2x + y^2 + 10 + 6y) = 4(x^2 + 4x + y^2 + 8 - 4y)\\[1em] ⇒9x^2 - 18x + 9y^2 + 90 + 54y = 4x^2 + 16x + 4y^2 + 32 - 16y\\[1em] ⇒9x^2 - 18x + 9y^2 + 90 + 54y - 4x^2 - 16x - 4y^2 - 32 + 16y = 0\\[1em] ⇒ 5x^2 + 5y^2 - 34x + 70y + 58 = 0\\[1em]

Hence, proved :- 5x2 + 5y2 - 34x + 70y + 58 = 0.

Question 18

The points A(3, 0), B(a, -2) and C(4, -1) are the vertices of triangle ABC right-angled at vertex A. Find the value of a.

Answer

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

The length of AC:

=(4−3)2+((−1)−0)2=12+(−1)2=1+1=2= \sqrt{(4 - 3)^2 + ((-1) - 0)^2}\\[1em] = \sqrt{1^2 + (-1)^2}\\[1em] = \sqrt{1 + 1}\\[1em] = \sqrt{2}

The length of BC:

=(a−4)2+((−2)−(−1))2=(a−4)2+(−1)2=a2+16−8a+1=a2−8a+17= \sqrt{(a - 4)^2 + ((-2) - (-1))^2}\\[1em] = \sqrt{(a - 4)^2 + (-1)^2}\\[1em] = \sqrt{a^2 + 16 - 8a + 1}\\[1em] = \sqrt{a^2 - 8a + 17}

The length of AB:

=(a−3)2+((−2)−0)2=(a−3)2+(−2)2=a2+9−6a+4=a2−6a+13= \sqrt{(a - 3)^2 + ((-2) - 0)^2}\\[1em] = \sqrt{(a - 3)^2 + (-2)^2}\\[1em] = \sqrt{a^2 + 9 - 6a + 4}\\[1em] = \sqrt{a^2 - 6a + 13}

Using Pythagoras theorem in triangle ABC,

BC2 = AB2 + AC2

⇒(2)2+(a2−6a+13)2=(a2−8a+17)2⇒2+a2−6a+13=a2−8a+17⇒a2−6a+15=a2−8a+17⇒−6a+15+8a−17=0⇒2a−2=0⇒2a=2⇒a=22⇒a=1⇒ (\sqrt2)^2 + (\sqrt{a^2 - 6a + 13})^2 = (\sqrt{a^2 - 8a + 17})^2\\[1em] ⇒ 2 + a^2 - 6a + 13 = a^2 - 8a + 17\\[1em] ⇒ a^2 - 6a + 15 = a^2 - 8a + 17 \\[1em] ⇒ - 6a + 15 + 8a - 17 = 0\\[1em] ⇒ 2a - 2 = 0\\[1em] ⇒ 2a = 2 \\[1em] ⇒ a = \dfrac{2}{2} \\[1em] ⇒ a = 1

Hence, the value of a = 1.

Question 19

If two vertices of an equilateral triangle be (0,0) and (3, 3\sqrt{3}), find the third vertex using distance formula.

Answer

Given,

A = (0, 0)

B =(3, 3\sqrt{3})

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Length of AB:

=(3−0)2+(3−0)2=(3)2+(3)2=9+3=12= \sqrt{(3 - 0)^2 + (\sqrt{3} - 0)^2} \\[1em] = \sqrt{(3)^2 + (\sqrt{3})^2} \\[1em] = \sqrt{9 + 3} \\[1em] = \sqrt{12}

Let the third vertex be C(x, y)

Since the triangle is equilateral, the distance from C to A and C to B must also be 12\sqrt{12}.

Length of AC = Length of AB

⇒(x−0)2+(y−0)2=12Squaring on both sides⇒x2+y2=12........(1)\Rightarrow \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{12} \\[1em] \text{Squaring on both sides} \\[1em] \Rightarrow x^2 + y^2 = 12 ........(1)

Length of BC = Length of AB

⇒(x−3)2+(y−3)2=12Squaring on both sides⇒(x−3)2+(y−3)2=12⇒x2−6x+9+y2−23y+3=12⇒x2+y2−6x−23y+12=12⇒x2+y2−6x−23y=0..........(2)Substitute equation (1) in (2)⇒12−6x−23y=0Divide by 2⇒6−3x−3y=0⇒3y=6−3x⇒y=6−3x3=23−x3⇒y=3(2−x)\Rightarrow \sqrt{(x - 3)^2 + (y - \sqrt{3})^2} = \sqrt{12} \\[1em] \text{Squaring on both sides} \\[1em] \Rightarrow (x - 3)^2 + (y - \sqrt{3})^2 = 12 \\[1em] \Rightarrow x^2 - 6x + 9 + y^2 - 2\sqrt{3}y + 3 = 12 \\[1em] \Rightarrow x^2 + y^2 - 6x - 2\sqrt{3}y + 12 = 12 \\[1em] \Rightarrow x^2 + y^2 - 6x - 2\sqrt{3}y = 0 ..........(2) \\[1em] \text{Substitute equation (1) in (2)} \\[1em] \Rightarrow 12 - 6x - 2\sqrt{3}y = 0 \\[1em] \text{Divide by 2} \\[1em] \Rightarrow 6 - 3x - \sqrt{3}y = 0 \\[1em] \Rightarrow \sqrt{3}y = 6 - 3x \\[1em] \Rightarrow y = \dfrac{6 - 3x}{\sqrt{3}} = 2\sqrt{3} - x\sqrt{3} \\[1em] \Rightarrow y = \sqrt{3}(2 - x)

Substitute the value of y in equation (1),

x2 + (3\sqrt{3}(2 - x))2 = 12

x2 + 3(4 - 4x + x2) = 12

x2 + 12 - 12x + 3x2 = 12

4x2 - 12x = 0

4x(x - 3) = 0

∴ x = 0 or x = 3

If x = 0, then

y = 3(2−0)=23\sqrt{3}(2 - 0) = 2\sqrt{3}

Vertex C = (0, 232\sqrt{3})

If x = 3, then

y = 3(2−3)=−3\sqrt{3}(2 - 3) = -\sqrt{3}

Vertex C = (3, −3-\sqrt{3}).

Hence, third vertex = (0, 232\sqrt{3}) and (3, −3-\sqrt{3}).

Case-Study Based Question

Question 1

A group of class IX students goes to a picnic during winter holidays. The positions of three friends Nitin, Rajesh and Kareem are shown by the points P, Q and R respectively.

A group of class IX students goes to a picnic during winter holidays. The positions of three friends Nitin, Rajesh and Kareem are shown by the points P, Q and R respectively. Distance Formula, Concise Mathematics Solutions ICSE Class 9.

(i) Find the distance between :

(a) Nitin and Rajesh

(b) Rajesh and Kareem

(c) Nitin and Kareem

(ii) Show that P, Q and R are collinear.

(iii) Find the point on x-axis which is equidistant from points Q and R.

(iv) If R is taken as origin; what will be the co-ordinates of P and Q ?

Answer

Given,

The co-ordinates are :

P(Nitin): (6, 4)

Q(Rajesh): (11, 9)

R(Kareem): (9, 7)

We know that

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

(i) (a) Distance between Nitin(P) and Rajesh(Q),

PQ=(11−6)2+(9−4)2=52+52=25+25=50=52.PQ = \sqrt{(11 - 6)^2 + (9 - 4)^2} \\[1em] = \sqrt{5^2 + 5^2} \\[1em] = \sqrt{25 + 25} \\[1em] = \sqrt{50} \\[1em] = 5\sqrt{2}.

Hence, distance between Nitin and Rajesh = 525\sqrt{2} units.

(b) Distance between Rajesh(Q) and Kareem(R),

QR=(11−9)2+(9−7)2=22+22=8=22.QR = \sqrt{(11 - 9)^2 + (9 - 7)^2} \\[1em] = \sqrt{2^2 + 2^2} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}.

Hence, distance between Rajesh and Kareem = 222\sqrt{2} units.

(c) Distance between Nitin(P) and Kareem(R),

PR=(9−6)2+(7−4)2=32+32=18=32.PR = \sqrt{(9 - 6)^2 + (7 - 4)^2} \\[1em] = \sqrt{3^2 + 3^2} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2}.

Hence, distance between Nitin and Kareem = 323\sqrt{2} units.

(ii) Points P, Q and R are collinear only if, PR + RQ = PQ

Consider L.H.S,

PR + RQ = 32+223\sqrt{2} + 2\sqrt{2}

= 525\sqrt{2}

Consider R.H.S,

PQ = 525\sqrt{2}

∴ L.H.S = R.H.S

So, they are collinear.

Hence, P, Q and R are collinear.

(iii) Let S (x, 0) be the point on x-axis equidistant from Q and R.

According to question,

SQ = SR

(x−11)2+(0−9)2=(x−9)2+(0−7)2\sqrt{(x - 11)^2 + (0 - 9)^2} = \sqrt{(x - 9)^2 + (0 - 7)^2}

Squaring on both sides,

(x - 11)2 + (0 - 9)2 = (x - 9)2 + (0 - 7)2

x2 - 22x + 121 + 81 = x2 - 18x + 81 + 49

-18x + 22x = 121 + 81 - 81 - 49

4x = 121 - 49

4x = 72

x = 724\dfrac{72}{4} = 18.

∴ The point S = (18, 0).

Hence, point on x-axis that is equidistant from Q and R = (18, 0).

(iv) If R is taken as origin, to find points of P and Q, their coordinates must be subtracted from the original coordinates of R(9, 7).

P' = (6 - 9, 4 - 7) = (-3, -3)

Q' = (11 - 9, 9 - 7) = (2, 2)

Hence, coordinates of P and Q are = (-3, -3) and (2, 2) respectively.

Question 2

Rohan and Sohan stay in a village which are 50 km apart and are situated on Delhi Agra Highway as shown in the figure. Another highway YY' crosses Agra Delhi Highway at O (0, 0). A small local road PQ crosses both the highways at points A and B such that OA = 10 km and OB = 12 km. Also the villages of Sita and Gita are on the smaller highway YY'. Sita's village is 12 km from O and Gita's village is 15 km from O.

Rohan and Sohan stay in a village which are 50 km apart and are situated on Delhi Agra Highway as shown in the figure. Another highway YY' crosses Agra Delhi Highway at O (0, 0). A small local road PQ crosses both the highways at points A and B such that OA = 10 km and OB = 12 km. Also the villages of Sita and Gita are on the smaller highway YY'. Sita's village is 12 km from O and Gita's village is 15 km from O. Distance Formula, Concise Mathematics Solutions ICSE Class 9.

Now answer the following questions:

(i) Find the Coordinates of A and B

(ii) Find the lengths of local road AB.

(iii) What is the shortest distance Sohan will cover to reach Sita's village.

(iv) How far is Rohan's village from Gita's village ?

Answer

(i) Point A is on the X-axis at a distance of 10 km from O (0, 0). So, A = (10, 0).

Point B is on the Y-axis at a distance of 12 km from O (0, 0). So, B = (0, 12).

Hence, coordinates of A = (10, 0) and coordinates of B = (0, 12).

(ii) Using Pythagoras theorem for the △OAB:

AB2 = BO2 + OA2

AB2 = 122 + 102

AB2 = 144 + 100

AB2 = 244

AB = 244\sqrt{244}

AB = 15.6 km.

Hence, length of the local road AB = 15.6 km.

(iii) Based on the diagram, Sohan's village is at (30, 0) and Sita's village is at (0, 12).

Distance=(x2−x1)2+(y2−y1)2=(30−0)2+(0−12)2=302+(−12)2=900+144=1044=32.31 km\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] = \sqrt{(30 - 0)^2 + (0 - 12)^2} \\[1em] = \sqrt{30^2 + (-12)^2} \\[1em] = \sqrt{900 + 144} \\[1em] = \sqrt{1044} \\[1em] = 32.31 \text{ km}

Hence, shortest distance between Sohan's village and Sita's village = 32.31 km.

(iv) Distance of Rohan's village from Gita's village:

Distance between Rohan's village to O + Distance between Gita's village to O

= 20 + 15

= 35 km.

Hence, distance from Rohan's village to Gita's village = 35 km.

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