Question

is point (x, y) is equidistant from point (7,1) and (3,5)show that y=x-2

Answer 1

Given

(x,y) is equidistant from (7,1) and (3,5).

From Distance formula,
${d}^{2} = {(x2 - x1) }^{2} + (y2 - y1)^{2}$
Distance of (x,y) from (7,1)=

${d}^{2} = {(x - 7)}^{2} + {(y - 1)}^{2}$
Distance of (x,y) from (3,5)=

${d}^{2} = {(x - 3)}^{2} + {(y - 5)}^{2}$

As the distances are equal,

${(x - 7)}^{2} + {(y - 1) }^{2} = {(x - 3) }^{2} + (y - 5)^{2}$

${x}^{2} - 14x + 49 + {y}^{2} - 2y + 1 = {x}^{2} - 6x + 9 + {y}^{2} - 10y + 25$

$- 14x + 49 - 2y + 1 + 6x + 10y - 25 - 9= 0 \\ - 8x + 8y = - 16 \\ - x + y = - 2 \\ \boxed{y = x - 2}$

Answer 2

halo mate your answer is here...



if you are satisfied then mark it brainlist answer