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

m (→‎Solution)
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== Solution ==
 
== Solution ==
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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]
 +
 +
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>
 
<math>\fbox{B}</math>
  

Revision as of 21:26, 20 December 2017

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).

$\fbox{B}$

See also

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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