# Class 9 SELINA Solutions Maths Chapter 28 - Distance Formula

## 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

AB^{2} = CD^{2}

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

9 + a^{2} + 4 + 4a = 9 + 9

a^{2} + 4a - 5 = 0

a^{2} - 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

PA^{2} = PB^{2}

(x - 5)^{2} + (y - 1)^{2} = (x - 11)^{2} + (y - 1)^{2}

x^{2} + 25 - 10x = x^{2} + 121 - 22x

12x = 96

x = 8

Also, PA = PC

PA^{2} = PC^{2}

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

x^{2} + 25 - 10x + y^{2} + 1 - 2y = x^{2} + 121 - 22x + y^{2} + 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

AB^{2} = 25

(0 - 3)^{2} + (y - 1 - 1)^{2} = 25

9 + y^{2} + 4 - 4y = 25

y^{2} - 4y - 12 = 0

y^{2} - 6y + 2y - 12 = 0

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

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

y = 6, -2

### Solution 19

AB = 17

AB^{2} = 289

(11 - x - 2)^{2} + (6 + 2)^{2} = 289

x^{2} + 81 - 18x + 64 = 289

x^{2} - 18x - 144 = 0

x^{2} - 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)

AB^{2} = 100

(-3 - 2x + 1)^{2} + (-1 - 3x - 1)^{2} = 100

(-2 - 2x)^{2} + (-2 - 3x)^{2} = 100

4 + 4x^{2} + 8x + 4 + 9x^{2} + 12x = 100

13x^{2} + 20x - 92 = 0

### Solution 21

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

PQ = 10

PQ^{2} = 100

(10 - 2)^{2} + (y + 3)^{2} = 100

64 + y^{2} + 9 + 6y = 100

y^{2} + 6y - 27 = 0

y^{2} + 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,

AT^{2} = PA^{2} - PT^{2} = 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:

AB^{2} + AC^{2} = BC^{2}

a^{2} - 6a + 13 + 2 = a^{2} - 8a + 17

2a = 2

a = 1