# SELINA Solutions for Class 9 Maths Chapter 28 - Distance Formula

Mathematics in ICSE Class 9 is one of the most challenging and trickiest subjects of all. It includes complex topics such as logarithms, expansions, indices and Pythagoras Theorem which are difficult to understand for an average student. TopperLearning provides study materials for ICSE Class 9 Mathematics to make the subject easy and help students to clear all their concepts. Our study materials comprise numerous video lessons, question banks, revision notes and sample papers which help achieve success in the examination.

Page / Exercise

## Chapter 28 - Distance Formula Exercise Ex. 28

Question 1

Find the distance between the following pairs of points:

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

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

(iii) and (iv) and 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 Question 2

Find the distance between the origin and the point:

(i) (-8, 6) (ii) (-5, -12) (iii) (8, -15)

Solution 2

Coordinates of origin are O (0, 0).

(i) A (-8, 6)

AO = (ii) B (-5, -12)

BO = (iii) C (8, -15)

CO = Question 3

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

Solution 3

It is given that the distance between the points A (3, 1) and B (0, x) is 5. Question 4

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

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).

Question 5

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

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).

Question 6

A point A is at a distance of unit from the point (4, 3). Find the co-ordinates of point A, if its ordinate is twice its abscissa.

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).

Question 7

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

Solution 7

Given that the point P (2, -1) is equidistant from the points A (a, 7) and B (-3, a). Question 8

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

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).

Question 9

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

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).

Question 10

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.

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 = Question 11

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

Solution 11 Since, PQ = QR, PQR is an isosceles triangle.

Question 12

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

Solution 12 Question 13

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.

Solution 13 Question 14

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

Solution 14 Since, AB = BC = CD = DA and AC = BD,

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

Question 15

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

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.

Question 16

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.

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.

Question 17

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

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).

Question 18

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

Question 19

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

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

Question 20

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 unit.

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 Question 21

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.

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).

Question 22

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

(i) AT (ii) AB. 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

Question 23

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

Solution 23 Question 24

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

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). Question 25

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

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). Question 26

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

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 .

Question 27

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.

Solution 27

It is given that PA: PB = 2: 3 Hence, proved.

Question 28

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.

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

## Key Features of Study Materials for ICSE Class 9 Maths:

• Include video lessons, question bank and solved sample papers
• Designed according to the latest ICSE syllabus
• Developed by subject experts
• Content revised from time to time
• Helpful when doing quick revision
• Guidance when doing homework
• Understand concepts easily
• Significantly improve your Mathematics score

CONTACT :
1800-212-78589:00am - 9:00pm IST all days.
SUBSCRIBE TO TOPPER :
Explore