# FINDING THE MISSING COORDINATE USING THE DISTANCE FORMULA

Finding the Missing Coordinate Using the Distance Formula :

In this section, we will learn, how to find the missing coordinate using the distance formula.

## Find the Missing Coordinate Using the Distance Formula Examples

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 PQ  =  10 units

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

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

=  √(10-2)2 + (y-(-3))2

√8² + (y + 3)² = 10

64 + y² + 6 y + 9 = 100

y² + 6 y + 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(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x.Also find the distances QR and PR.

Solution :

Distance between PQ  =  QR

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

Here x1  =  5, y1  =  -3, x2  =  0  and  y2  =  1

PQ   =   √(0-5)² + (1-(-3))²

=  √(2 - x)² + 25

Here x1  =  x, y1  =  0, x2  =  -2  and  y2  =  9

=  √(-2-x)² + (9-0)²

=  √(2 + x)² + (9)²

=  (2 + x)² + 81

√(2 - x)² + 25  =  (2+x)² + 81

4 + x² - 4 x + 25  =  4 + x² + 4 x + 81

x² - x² - 4 x - 4 x + 4 - 4  =  81 - 25

-8 x  =  56

x  =  -7

So, the required point is (-7, 0)

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)

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

√(x²-6x+9+y²-12y+36)  =  (x²+6x+9+y²-8y+16)

taking squares on both sides

xx+ y- y- 6x - 6x - 12y + 8y + 45 - 25  =  0

-12x - 4y + 20  =  0

divide the whole equation by (-4)

3 x + y - 5 = 0

After having gone through the stuff given above, we hope that the students would have understood, finding the missing coordinate using the distance formula.

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