**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.

**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 = √(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

A (–1, 1) and B (1, 3)

= √(1+1)^{2} + (3-1)^{2}

= √2^{2} + 2^{2}

= √(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 + a^{2} - 6a

8 = 13 + a^{2} - 6a

a^{2} - 6a + 13 - 8 = 0

a^{2} - 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)^{2 } = 10

Taking squares on both sides

(1-x)^{2} + (3-x)^{2 } = 100

1 - 2x + x^{2} + 9 - 6x + x^{2} = 100

2x^{2} - 8x + 10 - 100 = 0

2x^{2} - 8x - 90 = 0

Divide the entire equation by 2, we get

x^{2} - 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)^{2 } = √(-5-x)^{2} + (6-y)^{2 }

Taking squares on both sides

(3-x)^{2} + (4-y)^{2 } = (5+x)^{2} + (6-y)^{2 }

9 - 6x + x^{2} + 16 - 8y + y^{2} = 25 + 10x + x^{2} + 36 - 12y + y^{2}

-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 }

= √0^{2} + (-7)^{2}

= 7 ------(2)

AP = (3/7)AB

√(x - 2)^{2} + 9 = (3/7) 7

(x - 2)^{2} + 9 = 3^{2}

x^{2} - 4x + 4 + 9 - 9 = 0

x^{2} - 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)^{2 } = 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

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