The Selina Solutions for the questions given in Chapter 28, Distance Formula, of the Class 9 Selina textbooks are available here. In this chapter students learn about the topic of Distance Formula as well as the method of finding the distance between two points. Students can easily score full marks in the exams by solving all the questions present in the Selina textbook.
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Access Answers to Maths Selina Problems of Class 9 Chapter 28- Distance Formula
Exercise 28
1. Find the distance between the following pairs of points:
(i) (-3, 6) and (2, -6)
(ii) (-a, -b) and (a, b)
(iii) (3/5, 2) and 
(iv) (√3 + 1, 1) and (0, √3)
Solution:
(i) (-3, 6) and (2, -6)
Distance between the points is given by
= √[(2 -(-3))2 + (-6 – 6)2]
= √[52 + (-12)2]
= √(25 + 144)
= √169
= 13 units
Â
(ii) (-a, -b) and (a, b)
Distance between the points is given by
= √[(a -(-a))2 + (b – (-b))2]
= √[(2a)2 + (2b)2]
= √(4a2 + 4b2)
= √4(a2 + b2)
= 2√(a2 + b2) units
Â
(iii)Â (3/5, 2)Â andÂ
i.e., (3/5, 2) and (-1/5, 7/5)
Distance between the points is given by
= √[(-1/5 – 3/5)2 + (7/5 – 2)2]
= √[(-4/5)2 + ((7 – 10)/5)2]
= √(16/25 + 9/25)
= √(25/25)
= 1 unit
Â
(iv) (√3 + 1, 1) and (0, √3)
Distance between the points is given by
= √[(0 -(√3 + 1))2 + (√3 – 1)2]
= √[(√3 + 1)2 + (√3 – 1)2]
= √(3 + 1 + 2√3 + 3 + 1 – 2√3)
= √8
= 2√2 units
2. Find the distance between the origin and the point:
(i) (-8, 6) (ii) (-5, -12) (iii) (8, -15)
Solution:
Coordinates of the origin O are (0, 0)
Now, the distance between the origin and the points are
(i) A (-8, 6)
AO = √[(0 + 8)2 + (0 – 6)2]
= √[82 + (-6)2]
= √(64 + 36)
= √100
= 10 units
Â
(ii) B (-5, -12)
BO = √[(0 + 5)2 + (0 + 12)2]
= √[52 + 122]
= √(25 + 144)
= √169
= 13 units
Â
(iii) C (8, -15)
CO = √[(0 – 8)2 + (0 + 15)2]
= √[82 + 152]
= √(64 + 225)
= √289
= 17 units
3. The distance between the points (3, 1) and (0, x) is 5. Find x.
Solution:
Given, the distance between the points A (3, 1) and B (0, x) is 5
AB = 5
On squaring on both sides, we get
AB2 = 52
By distance formula, we have
(0 – 3)2 + (x – 1)2 = 25
9 + x2 – 2x + 1 = 25
x2 – 2x + 10 = 25
x2 – 2x – 15 = 0
On factorization, we get
x2 + 3x – 5x – 15 = 0
x(x + 3) – 5(x + 3) = 0
(x – 5) (x + 3) = 0
So, either (x – 5) = 0 or (x + 3) = 0
Hence,
x = 5 or -3
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:
Let’s assume the coordinates of the point on x-axis to be (x, 0)
Now, from the question, we have
√[(x – 11)2 + (0 + 8)2] = 17 [By using distance formula]
On squaring on both sides, we get
(x – 11)2 + (0 + 8)2 = 289
x2 + 121 – 22x + 64 = 289
x2 – 22x – 104 = 0
On factorization, we get
x2 – 26x + 4x – 104 = 0
x(x – 26) + 4(x – 26) = 0
(x – 26) (x + 4) = 0
So, either (x – 26) = 0 or (x + 4) = 0
Hence,
x = 26 or -4
Therefore, the required co-ordinates of the points on x-axis are (26, 0) and (-4, 0)
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:
Let’s assume the coordinates of the point on y-axis to be (0, y)
Now, from the question, we have
√[(0 + 8)2 + (y – 4)2] = 10 [By using distance formula]
On squaring on both sides, we get
82 + (y – 4)2 = 100
64 + y2 + 16 – 8y = 100
y2 – 8y – 20 = 0
On factorization, we get
y2 – 10y + 2y – 20 = 0
y(y – 10) + 2(y – 10) = 0
(y + 2) (y – 10) = 0
So, either (y + 2) = 0 or (y – 10) = 0
Hence,
y = 10 or -2
Therefore, the required co-ordinates of the points on y-axis are (0, 10) and (0, -2)
6. A point A is at a distance of √10 unit from the point (4, 3). Find the co-ordinates of point A, if its ordinate is twice its abscissa.
Solution:
Given, the co-ordinates of point A are such that its ordinate is twice its abscissa.
Now, let’s assume the co-ordinates of point A as (x, 2x)
And,
According to the question, we have
√[(x – 4)2 + (2x – 3)2] = √10 [By using distance formula]
On squaring on both sides, we get
(x – 4)2 + (2x – 3)2 = 10
x2 + 16 – 8x + 4x2 + 9 – 12x = 10
5x2 – 20x + 25 = 10
5x2 – 20x + 15 = 0
Dividing by 5, we get
x2 – 4x + 3 = 0
On factorization, we get
x2 – 3x – x + 3 = 0
x(x – 3) – (x – 3) = 0
(x – 3) (x – 1) = 0
So, either (x – 3) = 0 or (x -1) = 0
Hence,
x = 3 or 1
Thus, the co-ordinates of the point A are (1, 2) and (3, 6)
7. A point P (2, -1) is equidistant from the points (a, 7) and (-3, a). Find a.
Solution:
Given, the point P (2, -1) is equidistant from the points A (a, 7) and B (-3, a)
So, we have
PA = PB
⇒ PA2 = PB2
By using the distance formula, we have
(a – 2)2 + (7 + 1)2 = (-3 – 2)2 + (a + 1)2
a2 + 4 – 4a + 64 = 25 + a2 + 1 + 2a
68 – 4a = 26 + 2a
6a = 42
a = 7
Hence, the value of a is 7
8. What point on the x-axis is equidistant from the points (7, 6) and (-3, 4)?
Solution:
Let’s assume the co-ordinates of the required point on the x-axis to be P (x, 0)
The given points are A (7, 6) and B (-3, 4)
Given, PA = PB
So, on squaring on both sides, we get
PA2 = PB2
(x – 7)2 + (0 – 6)2 = (x + 3)2 + (0 – 4)2
x2 + 49 – 14x + 36 = x2 + 9 + 6x + 16
85 – 14x = 6x + 25
20x = 60
x = 3
Therefore, the required point is (3, 0)
9. Find a point on the y-axis which is equidistant from the points (5, 2) and (-4, 3).
Solution:
Let’s assume the co-ordinates of the required point on the y-axis to be P (0, y)
And. the given points are A (5, 2) and B (-4, 3)
Given, PA = PB
So, on squaring on both sides, we get
PA2 = PB2
(0 – 5)2 + (y – 2)2 = (0 + 4)2 + (y – 3)2
25 + y2 + 4 – 4y = 16 + y2 + 9 – 6y
29 – 4y = 25 – 6y
2y = -4
y = -2
Thus, the required point is (0, -2).
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:
(i) As the point P lies on the x-axis, its ordinate will be 0
(ii) As the point Q lies on the y-axis, its abscissa will be 0
(iii) The co-ordinates of P and Q are (-12, 0) and (0, -16) respectively
And,
PQ = √[(-12 – 0)2 + (0 + 16)2]
= √(144 + 256)
= √400
= 20
11. Show that the points P (0, 5), Q (5, 10) and R (6, 3) are the vertices of an isosceles triangle.
Solution:
Given points are P (0, 5), Q (5, 10) and R (6, 3)
Calculating:
PQ = √[(5 – 0)2 + (10 – 5)2]
= √(25 + 25)
= √50
= 5√2
QR = √[(6 – 5)2 + (3 – 10)2]
= √(1 + 49)
= √50
= 5√2
RP = √[(0 – 6)2 + (5 – 3)2]
= √(36 + 4)
= √40
= 2√10
As PQ = QR
Hence, ∆PQR is an isosceles triangle.
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:
Given points are P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1)
Calculating:
PQ = √[(6 – 0)2 + (2 + 4)2] = √(36 + 36) = √72 = 6√2 units
QR = √[(6 – 3)2 + (2 – 5)2] = √(9 + 9) = √18 = 3√2 units
RS = √[(3 + 3)2 + (5 + 1)2] = √(36 + 36) = √72 = 6√2 units
PS = √[(-3 – 0)2 + (-1 + 4)2] = √(9 + 9) = √18 = 3√2 units
PR = √[(3 – 0)2 + (5 + 4)2] = √(9 + 81) = √90 = 3√10 units
QS = √[(6 + 3)2 + (2 – 5)2] = √(9 + 9) = √18 = 3√2 units
It’s seen that,
PQ = RS and QR = PS
Also, PR = QS
Hence, PQRS is a rectangle
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:
Given points are A (1, -3), B (-3, 0) and C (4, 1)
Calculating:
AB = √[(-3 – 1)2 + (0 + 3)2] = √(16 + 9) = √25 = 5 units
BC = √[(4 + 3)2 + (1 – 0)2] = √(49 + 1) = √50 = 5√2 units
CA = √[(1 – 4)2 + (-3 – 1)2] = √(9 + 16) = √25 = 5 units
Hence, its seen that AB = CA
So, A, B and C are the vertices of an isosceles triangle
And,
AB2 + CA2 = 25 + 25 = 50
BC2 = (5√2)2 = 25 x 2 = 50
Thus, AB2 + CA2 = BC2
Therefore, we can conclude that A, B and C are the vertices of a right-angle triangle
i.e., ∆ABC is a right-angle isosceles triangle
Now,
Area of ∆ABC = ½ x AB x CA
= ½ x 5 x 5
= 25/2
= 12.5 sq. units
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:
Given points are A (5, 6), B (1, 5), C (2, 1) and D (6, 2)
Calculating the sides:
AB = √[(1 – 5)2 + (5 – 6)2] = √(16 + 1) = √17 units
BC = √[(2 – 1)2 + (1 – 5)2] = √(1 + 16) = √17 units
CD = √[(6 – 2)2 + (2 – 1)2] = √(16 + 1) = √17 units
DA = √[(5 – 6)2 + (6 – 2)2] = √(1+ 16) = √17 units
Now, calculating the diagonals:
AC = √[(2 – 5)2 + (1 – 6)2] = √(9 + 25) = √34 units
BD = √[(6 – 1)2 + (2 – 5)2] = √(25 + 9) = √34 units
As, AB = BC = CD = DA and AC = BD
Hence, we can conclude that A, B, C and D are the vertices of a square
15. Show that (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.
Solution:
Let the given points be taken as A (-3, 2), B (-5, -5), C (2, -3) and D (4, 4)
Calculating the sides:
AB = √[(-5 + 3)2 + (-5 – 2)2] = √(4 + 49) = √53 units
BC = √[(2 + 5)2 + (-3 + 5)2] = √(49 + 4) = √53 units
CD = √[(4 – 2)2 + (4 + 3)2] = √(4 + 49) = √53 units
DA = √[(-3 – 4)2 + (2 – 4)2] = √(49 + 4) = √53 units
Now, calculating the diagonals:
AC = √[(2 + 3)2 + (-3 – 2)2] = √(25 + 25) = √50 = 5√2 units
BD = √[(4 – 5)2 + (4 + 5)2] = √(81 + 81) = √162 = 9√2 units
As, AB = BC = CD = DA and AC ≠BD
Hence, we can conclude that the given vertices are of a rhombus
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:
Given, points A (-3, -2), B (-6, a), C (-3, -4) and D (0, -1) are the vertices of quadrilateral ABCD
Also given that AB = CD
On squaring on both sides, we get
AB2Â = CD2
(-6 + 3)2Â + (a + 2)2Â = (0 + 3)2Â + (-1 + 4)2 [By distance formula]
9 + a2Â + 4 + 4a = 9 + 9
a2Â + 4a – 5 = 0
a2Â – a + 5a – 5 = 0
a(a – 1) + 5 (a – 1) = 0
(a – 1) (a + 5) = 0
So, either (a – 1) = 0 or (a + 5) = 0
⇒ a = 1 or -5
Â
It is given that ‘a’ is negative,
Hence, the value of a is -5
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:
Let’s assume the circumcentre to be P (x, y)
Given, vertices of a triangle are (5, 1), (11, 1) and (11, 9)
If P is the circumcentre,
Then, PA = PB
Squaring on both sides, we get
PA2Â = PB2
(x – 5)2Â + (y – 1)2Â = (x – 11)2Â + (y – 1)2 [By distance formula]
x2Â + 25 – 10x = x2Â + 121 – 22x
12x = 96
x = 8
Â
Also, PA = PC
Squaring on both sides, we get
PA2Â = PC2
(x – 5)2Â + (y – 1)2Â = (x – 11)2Â + (y – 9)2
x2Â + 25 – 10x + y2Â + 1 – 2y = x2Â + 121 – 22x + y2Â + 81 – 18y
12x + 16y = 176
3x + 4y = 44
24 + 4y = 44
4y = 20
y = 5
Â
Therefore, the co-ordinates of the circumcentre of the triangle are (8, 5)
18. Given A = (3, 1) and B = (0, y – 1). Find y if AB = 5.
Solution:
Given, points A = (3, 1) and B = (0, y – 1)
According to the question, we have
AB = 5
On squaring it on both sides, we have
AB2Â = 25
(0 – 3)2Â + (y – 1 – 1)2Â = 25
9 + y2Â + 4 – 4y = 25
y2Â – 4y – 12 = 0
On factorization, we get
y2Â – 6y + 2y – 12 = 0
y(y – 6) + 2(y – 6) = 0
(y – 6) (y + 2) = 0
So, (y – 6) = 0 or (y + 2) = 0
Hence,
y = 6 or -2
19. Given A = (x + 2, -2) and B (11, 6). Find x if AB = 17.
Solution:
Given, points A = (x + 2, -2) and B (11, 6)
According to the question, we have
AB = 17
On squaring it on both sides, we have
AB2Â = 289
(11 – x – 2)2Â + (6 + 2)2Â = 289
x2Â + 81 – 18x + 64 = 289
x2Â – 18x – 144 = 0
x2Â – 24x + 6x – 144 = 0
x(x – 24) + 6(x – 24) = 0
(x – 24) (x + 6) = 0
So, (x – 24) = 0 or (x + 6) = 0
Hence,
x = 24, -6
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:
Given, the centre of the circle O is (2x – 1, 3x + 1)
Form the question, we have
Distance between the centre O (2x – 1, 3x + 1) and point A (-3, -1) should be equal to the radius of the circle
OA = 10 units (As given, diameter = 20 units)
On squaring on both sides, we get
OA2Â = 100
(-3 – 2x + 1)2Â + (-1 – 3x – 1)2Â = 100 [By distance formula]
(-2 – 2x)2Â + (-2 – 3x)2Â = 100
4 + 4x2Â + 8x + 4 + 9x2Â + 12x = 100
13x2Â + 20x – 92 = 0
By using the quadratic formula, we have
x = -20 ± √[(20)2 – 4(13)(-92)] / 26
= (-20 ± 72)/26
= -92/26 or 52/26
= -46/13 or 2
Therefore, the value of x is 2 or -46/13
Selina Solutions for Class 9 Maths Chapter 28- Distance Formula
Chapter 28, Distance Formula, consists of 1 exercise and the solutions given here contain answers to all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.
28.1 Introduction
28.2 The distance formula
28.3 Circumcentre of a triangle
Selina Solutions for Class 9 Maths Chapter 28- Distance Formula
For any two known points in a co-ordinate plane, the knowledge of co-ordinate geometry may be used to find:
- the distance between the given points
- the co-ordinates of a point which divides the line joining the given points in a given ratio
- the co-ordinates of the mid-point of the line segment joining the two given points,
- equation of the straight line through the given points
- equation of the perpendicular bisector of a line segment, etc.
Read and learn the Chapter 28 of Selina textbook to get to know more about Distance Formula. Learn the Selina Solutions for Class 9 effectively to attain excellent results in the examination.
