PROBLEMS BASED ON PROPERTIES OF DISTANCE IN COORDINATE GEOMETRY

About "Problems Based on Properties of Distance in Coordinate Geometry"

Problems Based on Properties of Distance in Coordinate Geometry :

Here we are going to see some example problems based on properties of distance in coordinate geometry.

Problems Based on Properties of Distance in Coordinate Geometry - Practice questions

Question 1 :

A (–1, 1), B (1, 3) and C (3, a) are points and if AB = BC, then find ‘a’.

Solution :

Distance between two points  =  √(x2 - x1)2 + (y2 - y1)2

A (–1, 1) and B (1, 3)

  =  √(1+1)2 + (3-1)2

  =  √22 + 22

  =  √(4+4)

AB   =  √8

B (1, 3) and C (3, a)

  =  √(1-3)2 + (3-a)2

  =  √(-2)2 + (3-a)2

BC  =  √4 + (3-a)2

AB  =  BC

√8  =  √4 + (3-a)2

Taking squares on both sides

8  =  4 + (3-a)2

8  =  4 + 9 + a2 - 6a

8  =  13 + a2 - 6a

a2 - 6a + 13 - 8  =  0

a2 - 6a + 5  =  0

(a - 1)(a - 5)  =  0

a  =  1 and a  =  5

Question 2 :

The abscissa of a point A is equal to its ordinate, and its distance from the point B(1, 3) is 10 units, What are the coordinates of A?

Solution :

Since the point A is equal to its ordinates

x  =  y

Let A(x, x)

Distance between the points A and B :

A(x, x)  B(1, 3)

√(1-x)2 + (3-x) =  10

Taking squares on both sides

(1-x)2 + (3-x) =  100

1 - 2x + x2 + 9 - 6x + x2  =  100

2x2 - 8x + 10 - 100  =  0

2x2 - 8x - 90  =  0

Divide the entire equation by 2, we get

x2 - 4x - 45  =  0

(x - 9)(x + 5)  =  0

x  =  9 and x  =  -5

A(9, 9) or B(-5, -5)

Question 3 :

The point (x, y) is equidistant from the points (3, 4) and

(–5,6). Find a relation between x and y.

Solution :

Let the points be p (x, y) A(3, 4) and B(-5, 6)

PA  =  PB

√(3-x)2 + (4-y) =  √(-5-x)2 + (6-y)

Taking squares on both sides

(3-x)2 + (4-y)2   =  (5+x)2 + (6-y)

9 - 6x + x2 + 16 - 8y + y2  =  25 + 10x + x2 + 36 - 12y + y2

-6x - 10x - 8y + 12y  + 25 - 61 =  0

-16x + 4y - 36 =  0

4x - y + 9  =  0

y  =  4x + 9

Question 4 :

Let A(2, 3) and B(2, –4) be two points. If P lies on the x-axis, such that AP = (3/7)AB, find the coordinates of P.

Solution :

Since the point P lies on x-axis, the y-coordinate will be 0.

A(2, 3) P(x, 0)

  =  √(x - 2)2 + (0 - 3)2  

  =  √(x - 2)2 + 9  ------(1)

A(2, 3) B(2, -4)

  =  √(2 - 2)2 + (-4 - 3)2  

  =  √02 + (-7)2 

  =  7 ------(2)

AP = (3/7)AB

√(x - 2)2 + 9   =  (3/7) 7

(x - 2)2 + 9   =  32

x2 - 4x + 4 + 9 - 9  =  0

x2 - 4x + 4  =  0

(x - 2) (x - 2)  =  0

x  =  2 and x  =  2

Hence the point P is (2, 0)

Question 5 :

Show that the point (11, 2) is the centre of the circle passing through the points (1, 2), (3, –4) and (5, –6)

Solution :

Let the center point be O (11, 2)

A(1, 2) B(3, -4) and C(5, -6)

AO  =  BO  =  CO  

O (11, 2) and A(1, 2)

=  √(1 - 11)2 + (1 - 11)2

=  √100 + 100

=  √200

O (11, 2) and B(3, -4)

=  √(3 - 11)2 + (-4 - 2)2

=  √(-8)2 + (-6)2

=  √64 + 36

=  √100

O (11, 2) and C (5, –6)

=  √(5 - 11)2 + (-6 - 2)2

=  √(-6)2 + (-8)2

=  √36 + 64

=  √100

Hence the given point (11, 2) is the center of the circle.

Question 6 :

The radius of a circle with centre at origin is 30 units. Write the coordinates of the points where the circle intersects the axes. Find the distance between any two such points.

Solution :

From the diagram given above, we come to know that the distance between the points (0, 0) and (a, 0) is 30 units.

√(0 - a)2 + (0 - 0) =  30

√(-a)2  =  30

a  =  30 

it is clearly shown that circle intersects the co-ordinate axes at four points. and that are (30,0) , (-30,0) ,(0,30) and (0,-30). 

now, distance between (30,0) and (0,30)

= √(30-0)² + (0 - 30)² = 30√2 unit 

Similarly , you can find distance between any such two points. 

for better understanding, let A = (30,0) , B=(0,30) , C = (-30,0) and D = (0, -30) 

then, Length of AB = length of BC = length of CD = length of DA = 30√2 unit

After having gone through the stuff given above, we hope that the students would have understood, "Problems Based on Properties of Distance in Coordinate Geometry" 

Apart from the stuff given in this section if you need any other stuff in math, please use our google custom search here.

Widget is loading comments...