ICSE Solutions for Selina Concise Chapter 28 Distance Formula Class 9 Maths
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 (-1/5 , 1 ²/5)
(iv) (√3+1, 1) and (0, √3)
Answer
(i) (-3 , 6) and (2, -6)
Distance between the given points
Distance between the given points
Distance between the given points
(i) (-8, 6)
(ii) (-5, -12)
(iii) (8, -15)
Answer
Coordinates of origin are O (0, 0).
(i) A (-8, 6) 
Answer
It is given that the distance between the points A(3, 1) and B(0, x) is 5.
∴ AB = 5
AB2 = 25
⇒ (0-3)2 + (x-1)2 = 25
⇒ 9 + x2 + 1 – 2x = 25
⇒ x2 – 2x – 15 = 15
⇒ x2 – 5x + 3x – 15 = 0
⇒ x(x-5) + 3(x-5) = 0
⇒ (x-5) (x+3) = 0
⇒ x = 5, -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).
Answer
Let the coordinates of the point on x - axis be (x,0).
From the given information, we have : 
⇒ x = 26 , -4
Thus, the required co - ordinates of the points on x -axis are (26, 0) and (-4, 0).
5 . Find the coordinates of the points on the y-axis, which are at a distance of 10 units from the point (-8, 4).
Answer
Let the coordinates of the point on y -axis be (0, y).
From the given information, we have : 
⇒ y = 10, -2
Thus, 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.
Answer
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 : 
⇒ x = 1, 3
Thus, the co -ordinates of the point A are (1, 2) and (3, 6).
Answer
Given that the point P (2, -1) is equidistant from the points A (a, 7) and B(-3, a).
∴ PA = PB
PA2 = PB2
⇒ (a-2)2 + (7+1)2 = (-3-2)2 + (a+1)2
⇒ a2 + 4 – 4a + 64 = 25 + a2 + 1 + 2a
⇒ 42 = 6a
⇒ a = 7
Answer
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
PA2 = PB2
⇒ (x-7)2 + (0-6)2 = (x+3)2 + (0-4)2
⇒ x2 + 49 – 14x + 36 = x2 + 9 + 6x + 16
⇒ 60 = 20 x
⇒ x = 3
Thus, 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).
Answer
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
PA2 = PB2
⇒ (0-5)2 + (y-2)2 = (0+4)2 + (y-3)2
⇒ 25 + y2 + 4 – 4y = 16 + y2 + 9 – 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.
Answer
(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. 
Answer
Answer
∵ PQ = RS and QR = PS,
Also PR = QS
∴ PQRS is a rectangle.
Answer
Hence, A, B, C are the vertices of a right - angled triangle.
Hence, △ABCD is an isosceles right - angled triangle.
Area of △ABC = (1/2) × AB ×CA
= (1/2) × 5×5
= 12.5 sq. units
Answer
Since, AB = BC = CD = DA and AC = BD,
A, B, C and D are the vertices of a square.
Answer
Since, AB = BC = CD = DA and AC ≠ BD
The given vertices are the vertices of a rhombus.
Answer
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.
Answer
Let the circumcntre 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
⇒ 12 x + 16 y = 176⇒ 3x + 4y = 44
⇒ 24 + 4y = 44
⇒ 4y = 20
⇒ y = 5
Thus, 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.
Answer
AB = 5
⇒ AB2 = 25
⇒ (0-3)2 + (y-1-1)2 = 25
⇒ 9 + y2 + 4 – 4y = 25
⇒ y2 -4y – 12 = 0
⇒ y2 – 6y + 6y - 12 = 0
⇒ y(y-6) + 2(y - 6) = 0⇒ (y - 6)(y +2) = 0
⇒ y = 6, -2
Answer
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
Answer
Distance between the points A (2x -1, 3x + 1) and B(-3 , -1) = Radius of circle
∴AB = 10(Since, diameter = 20 units, given)
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.
Answer
Let the co - ordinates of point Q be (10, y).
PQ = 10
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).
22. Point P (2, -7) is the center 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
Answer
(i) Given, radius = 13 units
∴ PA = PB = 13 units
Using distance formula, 
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
23. Calculate the distance between the points P (2, 2) and Q (5, 4) correct to three significant figures.
Answer
= 3.6055
= 3.61 units
Answer
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).
Answer
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).
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.
Answer
Let the required point of y - axis be P(0
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.
Answer
It is given that PA : PB = 2 : 3
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.
Answer
