FINDING THE MISSING COORDINATE USING THE DISTANCE FORMULA

Example 1 :

Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.

Solution :

Distance between P and Q  =  10 units.

Formula to find distance between two points :

√[(x2 - x1)2 + (y2 - y1)2]

Here x1 = 2, y1 = -3, x2 = 10  and  y2 = y.

Then, 

√[(10-2)2 + (y-(-3))2]  =  10

√[82 + (y + 3)2]  =  10

64 + y2 + 6y + 9  =  100

y2 + 6y + 9 + 64 - 100  =  0

y2 + 6 y - 27 = 0

(y + 9)(y - 3) = 0

 y + 9  =  0     y - 3  =  0

   y  =  -9     y  =  3

Example 2 :

If Q(x, 0) is equidistant from P(2, 5) and R(-2, 9), find the values of x. 

Solution :

From the given information,

PQ  =  QR -----(1)

Formula to find the distance between two points is 

√(x2 - x1)2 + (y2 - y1)2

To find PQ, substitute x1  =  5, y1  =  -3, x2  =  x, y2  =  0. 

PQ   =   √[(2-x)2 + (5-0)2]

PQ   =   √[(2-x)2 + 52]

PQ   =   √[(2-x)2 + 25]

To find QR, substitute x1  =  x, y1  =  0, x2  =  -2, y2  =  9.

RQ  =  √[(-2-x)2 + (9-0)2]

RQ  =  √[(2 + x)2 + (9)2]

  RQ  =  √[(2 + x)2 + 81]

(1)-----> PQ  =  RQ

√[(2-x)2 + 25]  =  √[(2 + x)2 + 81]

Square both sides. 

(2-x)2 + 25  =  (2+x)2 + 81

4 - 4x + x2 + 25  =  4 + 4x + x2 + 81

- 4x + 29  =  4x + 85

-56  =  8x

-7  =  x

Example 3 :

Find a relation between x and y such that the points (x, y) is equidistant from the points (3, 6) and (-3, 4).

Solution :

Point (x,y) is equidistant from the points (3, 6) and (-3, 4)

Then, 

√[(x-3)2 + (y-6)2]  =  √[(x+3)2 + (y-4)2]

Square both sides. 

(x-3)2 + (y-6)2  =  (x+3)2 + (y-4)2

x- 6x + 9 + y- 12y + 36  =  x+ 6x + 9 + y- 8y + 16

Simplify. 

- 6x + 9 - 12y + 36  =  6x + 9 - 8y + 16

- 6x - 12y + 45  =  6x - 8y + 25

-12x - 4y + 20  =  0

Divide both sides by -4. 

 3x + y - 5  =  0

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