There is a point P on the line $5x-3y=7$ that is equally far from the points $A(1,4) $ and $ B(3,10)$.
Find the coordinates of P.
What I did:
$5x-3y=7$ is the same as $ y = \dfrac{5}{3}x-2\dfrac{1}{3}$.
Line AB is $y=3x+1$ , and if we find the bisection of AB we can find point P because the bisection intersects (is it called that in English?) $ y = \dfrac{5}{3}x-2\dfrac{1}{3}$ at point P.
So we know 1 point on the bisection, which is $(3,7)$ . We also know the slope as $ 3 . - \dfrac{1}{3} = -1 $
So now we have can find that the formula of the bisection $ y = - \dfrac{1}{3}x + 8 $. Now we have to solve $ - \dfrac{1}{3}x + 8 = \dfrac{5}{3}x-2\dfrac{1}{3}$ and my answer is $5\dfrac{1}{6}$.
When we use substitution we find that $ y = 6\dfrac{5}{18}$
My question is whether this is the correct answer, since the coordinates are pretty 'messy', and if I did something wrong, what exactly?