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.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS