Question

If the point P(x, 3) is equidistant from the point A(7, −1) and B(6, 8), then find the value of x and find the distance AP.   

Answer 1

It is given that P(x, 3) is equidistant from the point A(7, −1) and B(6, 8).

∴ AP = BP

\[\Rightarrow \sqrt{\left( x - 7 \right)^2 + \left[ 3 - \left( - 1 \right) \right]^2} = \sqrt{\left( x - 6 \right)^2 + \left( 8 - 3 \right)^2}\]                  (Distance formula)

Squaring on both sides, we get

\[\left( x - 7 \right)^2 + 16 = \left( x - 6 \right)^2 + 25\]
\[ \Rightarrow x^2 - 14x + 49 + 16 = x^2 - 12x + 36 + 25\]
\[ \Rightarrow - 14x + 12x = 61 - 65\]
\[ \Rightarrow - 2x = - 4\]
\[ \Rightarrow x = 2\]

Thus, the value of x is 2.

\[\therefore AP = \sqrt{\left( 2 - 7 \right)^2 + \left[ 3 - \left( - 1 \right) \right]^2} = \sqrt{\left( - 5 \right)^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41}\] units