, 2019

The $y$-intercept of the line with equation $y=-2x+12$ is 12 and so $OA=12$.

The $x$-intercept of this line is determined by letting $y=0$ and solving for $x$. We get $0=-2x+12$ or $2x=12$ and so $x=6$.

The $x$-intercept is 6, and so $OB=6$.

The area of $\triangle AOB$ is $\dfrac12(OB)(OA)=\dfrac12(6)(12)=36$.

Solution 1

We begin by determining the equation of the line passing through $O$ and $C$.

This line is perpendicular to the line with equation$y=-2x+12$, and so its slope is the negative reciprocal of $-2$, which is $\dfrac12$.

This line passes through the origin and so it has $y$-intercept 0 and equation $y=\dfrac12x$.

Point $C$ is the point of intersection of the lines $y=\dfrac12x$ and $y=-2x+12$.

Substituting the equation of the first line into the second, we get $\dfrac12x=-2x+12$ or $\dfrac52x=12$ and so $x=\dfrac{24}{5}$.

When $x=\dfrac{24}{5}$, the equation $y=\dfrac12x$ gives $y=\dfrac12\left(\dfrac{24}{5}\right)=\dfrac{12}{5}$, and so the coodinates of $C$ are $\left(\dfrac{24}{5},\dfrac{12}{5}\right)$.

Solution 2

As in Solution 1, we begin by recognizing that the line passing through $O$ and $C$ has slope $\dfrac12$.

Point $C$ lies on the line with equation $y=-2x+12$ and so if the $x$-coordinate of $C$ is $a$, then the $y$-coordinate is $-2a+12$.

The slope of the line through $O(0,0)$ and $C(a,-2a+12)$ is $\dfrac{-2a+12}{a}$ and must equal $\dfrac12$.

Solving, we get $\dfrac{-2a+12}{a}=\dfrac12$ or $2(-2a+12)=a$ or $24=5a$, and so $a=\dfrac{24}{5}$.

When $a=\dfrac{24}{5}$, we get $-2a+12=-2\left(\dfrac{24}{5}\right)+12=-\dfrac{48}{5}+12=\dfrac{12}{5}$, and so the coordinates of $C$ are $\left(\dfrac{24}{5},\dfrac{12}{5}\right)$.
From part (b) Solution 1, the equation of the line passing through $O$ and $C$ is $y=\dfrac12x$. Point $D$ lies on this line and so if the $x$-coordinate of $D$ is $n$, then the $y$-coordinate of $D$ is $\dfrac12n$, so $D$ has coordinates $\left(n, \dfrac12n\right)$.

Point $E$ lies vertically below $D$ and thus has the same $x$-coordinate as $D$.

That is, the coordinates of $E$ are $(n,0)$ and so $OE=n$.

Similarly, $F$ is positioned horizontally from $D$ and thus has the same $y$-coordinate as $D$.

That is, the coordinates of $F$ are $\left(0,\dfrac12n\right)$ and so $OF=\dfrac12n$.

The area of $DEOF$ is 1352, and so $(OE)(OF)=1352$ or $n\left(\dfrac12n\right)=1352$ or $n^2=2704$, and so $n=\sqrt{2704}=52$ (since $\text{n gt;0}$), and $\dfrac12n=26$.

If the area of $DEOF$ is 1352, the coordinates of $D$ are $(52,26)$.