Problem 426

Let $A$ and $B$ be the points at which the line with
equation $y=mx+7$ intersects the
lines with equations $y=0$ and $y=2$, respectively. Also, suppose $C$ has coordinates $(0,2)$ and $D$ has coordinates $(0,0)$. The trapezoid in the problem is
$ABCD$, as shown.

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We can find the coordinates of $A$
and $B$ in terms of $m$.

To find the coordinates of $A$, we
find the point of intersection of the line with equation $y=mx+7$ and the line with equation $y=0$. Setting $0=mx+7$, we get $x=-\dfrac{7}{m}$. Note that the $y$-coordinate of $A$ must be $0$ since it is, by definition, on the line
with equation $y=0$.

Therefore, the coordinates of $A$ are
$\left(-\dfrac{7}{m},0\right)$.
Similarly, the coordinates of $B$ are
$\left(-\dfrac{5}{m},2\right)$.

Trapezoid $ABCD$ has parallel bases
$AD$ and $BC$ and height $CD$.

The two bases are horizontal and have lengths $AD=\dfrac{7}{m}$ and $BC=\dfrac{5}{m}$. The length of $CD$ is $2$.

Therefore, the area of $ABCD$ is
$\$\dfrac{1}{2}\cdot CD\cdot(AD+BC)=\dfrac{1}{2}\cdot 2\cdot\left(\dfrac{7}{m}+\dfrac{5}{m}\right)=\dfrac{12}{m}\$.$

It is given that the area of the trapezoid is $3$, so we have $\dfrac{12}{m}=3$, or $m=4$.