1964 AHSME Problems/Problem 15

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Problem 15

A line through the point $(-a,0)$ cuts from the second quadrant a triangular region with area $T$. The equation of the line is:

$\textbf{(A)}\ 2Tx+a^2y+2aT=0 \qquad \textbf{(B)}\ 2Tx-a^2y+2aT=0 \qquad \textbf{(C)}\ 2Tx+a^2y-2aT=0 \qquad \\ \textbf{(D)}\ 2Tx-a^2y-2aT=0 \qquad \textbf{(E)}\ \text{none of these}$

Solution

The right triangle has area $T$ and base $a$, so the height $h$ satisfies $\frac{1}{2}ha = T$. This means $h = \frac{2T}{a}$. Becuase the triangle is in the second quadrant, the coordinates are the origin, $(-a, 0)$, and $(0, h)$, which means the third coordinate is $(0, \frac{2T}{a})$.

So, we want the equation of a line though $(-a, 0)$ and $(0, \frac{2T}{a})$. The slope is $\frac{\frac{2T}{a} - 0}{0 - (-a)}$, which simplifies to $\frac{2T}{a^2}$.

The y-intercept is $(0, \frac{2T}{a})$, so the line in slope-intercept form is $y = mx + b$, or:


$y = \frac{2T}{a^2}x + \frac{2T}{a}$

All the solutions have a positive x-coefficient and no fractions, so we clear the fractions by multiplying by $a^2$ and move the $y$ term to the right to get:

$0 = 2Tx + 2aT - a^2y$

This is equivalent to option $\boxed{\textbf{(B)}}$

See Also

1964 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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