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

Revision as of 22:30, 20 December 2017

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

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

$\fbox{B}$

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 25	Followed by Problem 27
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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 == Solution ==
-The graph of <math>(x + y - 5)(2x - 3y + 5) = 0</math> is the union of the lines <math>x + y - 5 = 0</math> and <math>2x - 3y + 5 = 0</math> (shown in red below). The graph of <math>(x - y + 1)(3x + 2y - 12) = 0</math> is the union of the lines <math>x - y + 1 = 0</math> and <math>3x - 2y - 12 = 0</math> (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("<math>x</math>", (5,0), NE);
-label("<math>y</math>", (0,5), NE);
-dot("<math>(2,3)</math>", (2,3), E);
-label("<math>x + y - 5 = 0</math>", (5,0), E, red);
-label("<math>2x - 3y + 5 = 0</math>", (5,5), E, red);
-label("<math>x - y + 1 = 0</math>", (4,5), N, blue);
-label("<math>3x + 2y - 12 = 0</math>", (5,-3/2), E, blue);
-[/asy]
-https://latex.artofproblemsolving.com/c/w/o/cwocgeak.png?time=1513823210175
- Every line passes through the point <math>(2,3)</math>, so the intersection of the two graphs consists of exactly <math>\boxed{1}</math> point. The answer is (B).
 <math>\fbox{B}</math>