Question

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

Answer 1

Answer:

3x + y = 5

Step-by-step explanation:

Let P(x , y) , Q( 3, 6) and T( -3 , 4)

A/C to question ,

PQ = PT

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

take square both sides,

(x-3)² + ( y -6)² = (x +3)² + (y-4)²

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

-12x -4y + 20 = 0

3x + y = 5