1964 AHSME Problems/Problem 15

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 $\boxed{\textbf{(B)}}$

1964 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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1964 AHSME Problems/Problem 15

Problem 15

Solution

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