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

(ii) (-a, -b) and (a, b) 
Distance between the given points 


(iv) (√3 + 1, 1) and (0, √3)
Distance between the given points 


2. Find the distance between the origin and the point:

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

Answer

Coordinates of origin are O (0, 0). 
(i) A (-8, 6) 


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

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). 


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

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 


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

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. 


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

Answer 


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

Answer 


∵ PQ = RS and QR = PS, 
Also PR = QS 
∴ 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.

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


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.

Answer 


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

Answer


Since, AB = BC = CD = DA and AC ≠ BD
The given vertices are the vertices 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.

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. 


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

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 


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

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 


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.

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 


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

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).


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

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

Thus, the required points on  y - axis are (0, 9) and (0, 35/3). 

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 

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