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1955 AHSME Problems/Problem 46

The graphs of $2x+3y-6=0, 4x-3y-6=0, x=2$, and $y=\frac{2}{3}$ intersect in:

$\textbf{(A)}\ \text{6 points}\qquad\textbf{(B)}\ \text{1 point}\qquad\textbf{(C)}\ \text{2 points}\qquad\textbf{(D)}\ \text{no points}\\ \textbf{(E)}\ \text{an unlimited number of points}$

Solution

We first convert each of the lines into slope-intercept form ($y = mx + b$):

$2x+3y-6=0 ==> 3y = -2x + 6 ==> y = -\frac{2}{3}x + 2$

$4x - 3y - 6 = 0 ==> 4x - 6 = 3y ==> y = \frac{4}{3}x - 2$

$x = 2$ stays as is.

$y = \frac{2}{3}$ stays as is

We can graph the four lines here:[1]

When we do that, the answer turns out to be $\boxed{\textbf{(B)} \text{1 point}}$.

See Also

Go back to the rest of the 1955 AHSME Problems.

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC logo.png

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