SELINA Solutions for Class 9 Maths Chapter 28 - Distance Formula

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Chapter 28 - Distance Formula Exercise Ex. 28

Solution 1

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

Distance between the given points

 

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

Distance between the given points

 

(iii) and

Distance between the given points

 

(iv) and

Distance between the given points

Solution 2

Coordinates of origin are O (0, 0).

(i) A (-8, 6)

AO =

 

(ii) B (-5, -12)

BO =

 

(iii) C (8, -15)

CO =

Solution 3

It is given that the distance between the points A (3, 1) and B (0, x) is 5.

Solution 4

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

From the given information, we have:

Thus, the required co-ordinates of the points on x-axis are (26, 0) and (-4, 0).

Solution 5

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

From the given information, we have:

Thus, the required co-ordinates of the points on y-axis are (0, 10) and (0, -2).

Solution 6

It is given that the co-ordinates of point A are such that its ordinate is twice its abscissa.

So, let the co-ordinates of point A be (x, 2x).

We have:

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

Solution 7

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

Solution 8

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

The given points are A (7, 6) and B (-3, 4).

Given, PA = PB

Thus, the required point is (3, 0).

Solution 9

Let the co-ordinates of the required point on y-axis be P (0, y).

The given points are A (5, 2) and B (-4, 3).

Given, PA = PB

Thus, the required point is (0, -2).

Solution 10

(i) Since, the point P lies on the x-axis, its ordinate is 0.

(ii) Since, the point Q lies on the y-axis, its abscissa is 0.

(iii) The co-ordinates of P and Q are (-12, 0) and (0, -16) respectively.

 

PQ =

Solution 11

Since, PQ = QR, PQR is an isosceles triangle.

Solution 12

Solution 13

Solution 14

Since, AB = BC = CD = DA and AC = BD,

A, B, C and D are the vertices of a square.

Solution 15

Let the given points be A (-3, 2), B (-5, -5), C (2, -3) and D (4, 4).

 

Since, AB = BC = CD = DA and AC BD

The given vertices are the vertices of a rhombus.

Solution 16

AB = CD

AB2 = CD2

(-6 + 3)2 + (a + 2)2 = (0 + 3)2 + (-1 + 4)2

9 + a2 + 4 + 4a = 9 + 9

a2 + 4a - 5 = 0

a2 - a + 5a - 5 = 0

a(a - 1) + 5 (a - 1) = 0

(a - 1) (a + 5) = 0

a = 1 or -5

 

It is given that a is negative, thus the value of a is -5.

Solution 17

Let the circumcentre be P (x, y).

Then, PA = PB

PA2 = PB2

(x - 5)2 + (y - 1)2 = (x - 11)2 + (y - 1)2

x2 + 25 - 10x = x2 + 121 - 22x

12x = 96

x = 8

 

Also, PA = PC

PA2 = PC2

(x - 5)2 + (y - 1)2 = (x - 11)2 + (y - 9)2

x2 + 25 - 10x + y2 + 1 - 2y = x2 + 121 - 22x + y2 + 81 - 18y

12x + 16y = 176

3x + 4y = 44

24 + 4y = 44

4y = 20

y = 5

 

Thus, the co-ordinates of the circumcentre of the triangle are (8, 5).

Solution 18

AB = 5

AB2 = 25

(0 - 3)2 + (y - 1 - 1)2 = 25

9 + y2 + 4 - 4y = 25

y2 - 4y - 12 = 0

 

y2 - 6y + 2y - 12 = 0

y(y - 6) + 2(y - 6) = 0

(y - 6) (y + 2) = 0

y = 6, -2

Solution 19

AB = 17

AB2 = 289

(11 - x - 2)2 + (6 + 2)2 = 289

x2 + 81 - 18x + 64 = 289

x2 - 18x - 144 = 0

x2 - 24x + 6x - 144 = 0

x(x - 24) + 6(x - 24) = 0

(x - 24) (x + 6) = 0

x = 24, -6

Solution 20

Distance between the points A (2x - 1, 3x + 1) and B (-3, -1) = Radius of circle

AB = 10 (Since, diameter = 20 units, given)

AB2 = 100

(-3 - 2x + 1)2 + (-1 - 3x - 1)2 = 100

(-2 - 2x)2 + (-2 - 3x)2 = 100

4 + 4x2 + 8x + 4 + 9x2 + 12x = 100

13x2 + 20x - 92 = 0

Solution 21

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

PQ = 10

PQ2 = 100

(10 - 2)2 + (y + 3)2 = 100

64 + y2 + 9 + 6y = 100

y2 + 6y - 27 = 0

y2 + 9y - 3y - 27 = 0

y(y + 9) - 3(y + 9) = 0

(y + 9) (y - 3) = 0

y = -9, 3

 

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

Solution 22

(i) Given, radius = 13 units

PA = PB = 13 units

 

Using distance formula,

 

Using Pythagoras theorem in PAT,

AT2 = PA2 - PT2 = 169 - 25 = 144

AT = 12 units

 

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

AB = 2AT = 2 12 units = 24 units

Solution 23

Solution 24

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

Abscissa of point B = 11

Since, B lies of x-axis, so its co-ordinates are (11, 0).

Solution 25

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

Ordinate of point B = 9

Since, B lies of y-axis, so its co-ordinates are (0, 9).

Solution 26

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

 

From the given information, we have:

 

Thus, the required points on y-axis are (0, 9) and.

Solution 27

It is given that PA: PB = 2: 3

Hence, proved.

Solution 28

Since, triangle ABC is a right-angled at A, we have:

AB2 + AC2 = BC2

a2 - 6a + 13 + 2 = a2 - 8a + 17

2a = 2

a = 1