Question

The point on x-axis which is equidistant from (6,-3) and (10,8)
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Answer 1

Answer:

as given the point which is equidistant is on x axis so the coordinate is (x,0)

let it be P

let the distant between the points AP and BP

√(6-x)²+(-3-0)² =√(10-x)² +(8-0)²

remove roots by taking any one of them to other side and then cut

root by square

then our eq. becomes

(6-x)²+9 = (10-x)² + 64

now solve brackets

36+x²-12x+9 = 100 +x²-20x+64

now solve for x

the eq. reduce to

8x = 164-45

x = 119/8

hope its helpful!!