# Difference between revisions of "1970 AHSME Problems/Problem 15"

## Problem

Lines in the $xy$-plane are drawn through the point $(3,4)$ and the trisection points of the line segment joining the points $(-4,5)$ and $(5,-1)$. One of these lines has the equation

$\text{(A) } 3x-2y-1=0\quad \text{(B) } 4x-5y+8=0\quad \text{(C) } 5x+2y-23=0\quad\\ \text{(D) } x+7y-31=0\quad \text{(E) } x-4y+13=0$

## Solution

The trisection points of $(-4, 5)$ and $(5, -1)$ can be found by trisecting the x-coordinates and the y-coordinates separately. The difference of the x-coordinates is $9$, so the trisection points happen at $-4 + \frac{9}{3}$ and $-4 + \frac{9}{3} + \frac{9}{3}$, which are $-1$ and $2$. Similarly, the y-coordinates have a difference of $6$, so the trisections happen at $3$ and $1$. So, the two points are $(-1, 3)$ and $(2, 1)$.

We now check which line has both $(3, 4)$ and one of the two trisection points on it. Plugging in $(x, y) = (3, 4)$ into all five of the equations works. The point $(2, 1)$ doesn't work in any of the five lines. However, $(-1, 3)$ works in line $E$. Thus, the answer is $\fbox{E}$