# Difference between revisions of "1970 AHSME Problems/Problem 26"

## Problem

The number of distinct points in the $xy$-plane common to the graphs of $(x+y-5)(2x-3y+5)=0$ and $(x-y+1)(3x+2y-12)=0$ is

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) } 4\quad \text{(F) } \infty$

## Solution

The graph of $(x + y - 5)(2x - 3y + 5) = 0$ is the union of the lines $x + y - 5 = 0$ and $2x - 3y + 5 = 0$ (shown in red below). The graph of $(x - y + 1)(3x + 2y - 12) = 0$ is the union of the lines $x - y + 1 = 0$ and $3x - 2y - 12 = 0$ (shown in blue below).

[asy] unitsize(0.8 cm);

draw((-1,0)--(5,0)); draw((0,-1)--(0,5));

draw((0,5)--(5,0),linewidth(1)+red); draw((-1,1)--(5,5),linewidth(1)+red); draw((-1,0)--(4,5),linewidth(1)+blue); draw((2/3,5)--(5,-3/2),linewidth(1)+blue);

label("$x$", (5,0), NE); label("$y$", (0,5), NE);

dot("$(2,3)$", (2,3), E); label("$x + y - 5 = 0$", (5,0), E, red); label("$2x - 3y + 5 = 0$", (5,5), E, red); label("$x - y + 1 = 0$", (4,5), N, blue); label("$3x + 2y - 12 = 0$", (5,-3/2), E, blue); [/asy]

Every line passes through the point $(2,3)$, so the intersection of the two graphs consists of exactly $\boxed{1}$ point. The answer is (B).


$\fbox{B}$