, 2021

Trapezoid $ABCD$ is drawn, as shown.

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The slope of line segments $AB$ and $CD$ are each zero and thus they are parallel.

The length of $AB$ is the difference between the $x$-coordinates of $A$ and $B$, or 12.
The length of $CD$ is the difference between the $x$-coordinates of $C$ and $D$, or $11-2=9$.

The height of the trapezoid is equal to the vertical distance between $AB$ and $CD$, which is 5.
The area of trapezoid $ABCD$ is $\dfrac{5}{2}(AB+CD)$ or $\dfrac{5}{2}(21)=\dfrac{105}{2}$.
The line passing through $B$ and $D$ intersects the $y$-axis at $E$. Let the coordinates of $E$ be $(0,e)$, as shown.

The slope of the line through $B$ and $D$ is $\dfrac{5-0}{2-12}=-\dfrac12$.

Solution 1

Since $E,D$ and $B$ lie on the same line, then the slope of $ED$ is equal to the slope of $BD$.

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Equating slopes, we get $\dfrac{e-5}{0-2}=-\dfrac12$ or $\dfrac{e-5}{2}=\dfrac12$, and so $e-5=1$ or $e=6$.

Thus, point $E$ has coordinates $(0,6)$.

Solution 2

The line passing through $B(12,0)$ and $D(2,5)$ has slope $-\dfrac12$, and thus has equation $y-5=-\dfrac12(x-2)$.

Rearranging, we get $y-5=-\dfrac12x+1$ or $y=-\dfrac12x+6$.

Since this line has $y$-intercept 6, then point $E$ has coordinates $(0,6)$.

Solution 3

The line passing through $B(12,0)$ and $D(2,5)$ has slope $-\dfrac12$, and thus has equation $y-5=-\dfrac12(x-2)$.

This line passes through $E(0,e)$, and so $e-5=-\dfrac12(0-2)$ or $e=1+5=6$.
Thus, point $E$ has coordinates $(0,6)$.
Sides $AD$ and $BC$ are extended to intersect at $F$, as shown.

Solution 1

Let the coordinates of $F$ be $(j,k)$.

Since $A,D$ and $F$ lie on the same line, then the slope of $AD$ is equal to the slope of $AF$.

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Equating slopes, we get $\dfrac{5}{2}=\dfrac{k}{j}$ or $k=\dfrac{5}{2}j$.
Since $B,C$ and $F$ lie on the same line, then the slope of $BC$ is equal to the slope of $BF$.

Equating slopes, we get $\dfrac{5-0}{11-12}=\dfrac{k-0}{j-12}$ or $-5=\dfrac{k}{j-12}$, and so $k=-5(j-12)$.

Substituting $k=\dfrac{5}{2}j$, we get $\dfrac{5}{2}j=-5(j-12)$ or $j=-2(j-12)$, and so $3j=24$ or $j=8$.

When $j=8$, $k=\dfrac52(8)=20$, and so $F$ has coordinates $(8,20)$.

Solution 2

The line passing through $A(0,0)$ and $D(2,5)$ has slope $\dfrac52$ and $y$-intercept 0, and thus has equation $y=\dfrac52x$.

The line passing through $B(12,0)$ and $C(11,5)$ has slope $-5$ and thus has equation $y=-5(x-12)$.

These two lines intersect at $F$, and so the coordinates of $F$ can be determined by solving the equation $\dfrac52x=-5(x-12)$. Solving, we get $x=-2(x-12)$ or $3x=24$, and so $x=8$.
When $x=8$, $y=\dfrac52(8)=20$, and so $F$ has coordinates $(8,20)$.
Let $P$ have coordinates $(r,s)$.

Assume $AB$ is the base of $\triangle PAB$.

In this case, if the height of $\triangle PAB$ is $h$, then the area of $\triangle PAB$ is $\dfrac12(AB)h=6h$.

The area of $\triangle PAB$ is 42, and so $6h=42$ or $h=7$.

That is, $P(r,s)$ is located a vertical distance of 7 units from the line through $A$ and $B$, or 7 units from the $x$-axis.

There are two possibilities: $P(r,s)$ is located 7 units above the $x$-axis, and thus lies on the horizontal line $y=7$, or $P(r,s)$ is located 7 units below the $x$-axis, and thus lies on the horizontal line $y=-7$.

In the first case, $P$ has coordinates $(r,7)$ and in the second case, $P$ has coordinates $(r,-7)$.

Recall that $P$ lies on the line passing through $B$ and $D$.

The line passing through $B(12,0)$ and $D(2,5)$ has slope $-\dfrac12$, and thus has equation $y-5=-\dfrac12(x-2)$.

If $P(r,7)$ lies on this line, then $7-5=-\dfrac12(r-2)$ or $-4=r-2$, and so $r=-2$.

Similarly, if $P(r,-7)$ lies on this line, then $-7-5=-\dfrac12(r-2)$ or $24=r-2$, and so in this case, $r=26$.
The points $P$ that lie on the line passing through $B$ and $D$, so that the area of $\triangle PAB$ is 42, are $(-2,7)$ and $(26,-7)$.