Question

Find a relation between x and y such that the point (x,y) is equidistant from the points (7,1) and (3,5)

Answer 1

Hii friend,

Let A(7,1) , B(3,5) are the given points.

We have,

PA = PB => PA² = PB²

(X-7)² + (1-Y)² = (3-x)² + (5-Y)²

X² + (7)² - 2 × X × 7 + (1)² + (Y)² - 2 × 1 × Y = (3)² + (X)² -2 × 3 × X + (5)² + (Y)² - 2 × 5 × Y.

X²+49 -14X + 1 +Y²- 2Y = 9 + X² - 6X + 25 + Y² - 10Y.

X²+Y² -14X-2Y +50 = X²+Y² -6X +10Y +34

X²-X²+Y²-Y²-14X+6X-2Y-10Y = 34-50

-8X -12Y = -16

8X+12Y = 16

4(2X+3Y) = 16

2X +3Y = 16/4

2X+3Y = 4

2X = 4-3Y

X = 4-3Y/2

Hence,

X = 4-3Y/2 is the desired relation between X and Y.


HOPE IT WILL HELP YOU...... :-)