SOLUTION: Find the point (𝑥,𝑦) on the line 𝑦=𝑥 that is equidistant from the points (10,−10) and (−4,6) .

Algebra ->  Linear-equations -> SOLUTION: Find the point (𝑥,𝑦) on the line 𝑦=𝑥 that is equidistant from the points (10,−10) and (−4,6) .      Log On


   

Question 1203786: Find the point (𝑥,𝑦)
on the line 𝑦=𝑥
that is equidistant from the points (10,−10)
and (−4,6)
.
Found 3 solutions by josgarithmetic, mananth, ikleyn:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Because you want x=y or y=x


sqrt%28%28x-10%29%5E2%2B%28y%2B10%29%5E2%29=sqrt%28%28x%2B4%29%5E2%2B%28y-6%29%5E2%29
sq. both sides, and making substitution
%28x-10%29%5E2%2B%28x%2B10%29%5E2=%28x%2B4%29%5E2%2B%28x-6%29%5E2
.
.
.
The point, (-37,-37)

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Find the point (𝑥,𝑦)
on the line 𝑦=𝑥
that is equidistant from the pointsA (10,−10)
and B(−4,6)
The distance between two points (x₁, y₁) and (x₂, y₂) is given by:
Distance+=+sqrt%28%28x1-x2%29%5E2+%2B+%28y1+-+y2%29%5E2%29
Square both sides andplug the given values
(x+4)^2+(y-6)^2= (x-10)^2+(y+10)^2
expand
x^2+8x+16 +y^2-12y+36 = x^2-20x+100 +y^2+20y+100

8x+16-12y+36 = -20x +100+20y+100
combine like terms
28x-32y =148 but x=y
-4x = 148
x=37 ---> y=37
(37,37)
point (x, y) that is equidistant from the points (10, -10) and (-4, 6) is indeed (37, 37).

Answer by ikleyn(52790) About Me  (Show Source):
You can put this solution on YOUR website!
.

The answer  (37,37)  by @mananth is   INCORRECT.