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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?

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The bisection point is $(2,7)$. Also which line has slope $-1$?The logic is right.but the calculations are wrong. –  Phani Raj Oct 6 '12 at 14:51

1 Answer 1

up vote 1 down vote accepted

The equation of the line passes the midpoint $M(2,7)$ of the segment $AB$ and perpendicular to the line $AB$ is $x+3y-23 = 0$. The coordinates of the needing point is the solution of the system of equations $$x+3y-23 = 0, \quad 5x-3y=7.$$ And then, we have $(5,6)$.

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Exactly. Thank you! –  JohnPhteven Oct 6 '12 at 15:17

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