The line $y=2 x+4$ intersects the $y$-axis at $R$, as shown. A second line, parallel to the $y$-axis, is drawn through $P(p, 0)$, with $p>0$. These two lines intersect at $Q$.

(a) Determine the length of $O R$. (Note that $O$ is the origin $(0,0)$.)

(b) Determine the coordinates of point $Q$ in terms of $p$.

(c) If $p=8$, determine the area of $O P Q R$.

(d) If the area of $O P Q R$ is 77, determine the value of $p$.

![](https://cdn.mathpix.com/cropped/2024_04_17_baced599e3620c07808bg-3.jpg?height=474&width=439&top_left_y=1015&top_left_x=1190)
(a) The $y$-intercept of the line with equation $y=2 x+4$ is 4 .

Since $R$ is the point where this line crosses the $y$-axis, then $R$ has coordinates $(0,4)$. Since $O$ has coordinates $(0,0)$, the length of $O R$ is 4 .

(b) The point $Q$ is the point of intersection of the line with equation $y=2 x+4$ and the vertical line with equation $x=p$.

Thus, the $x$-coordinate of $Q$ is $p$, and its $y$-coordinate is thus $2 p+4$.

So the coordinates of $Q$ are $(p, 2 p+4)$.

(c) When $p=8$, the coordinates of $P$ are $(8,0)$ and the coordinates of $Q$ are $(8,20)$.

Quadrilateral $O P Q R$ is a trapezoid with parallel sides $O R$ and $P Q$, and height equal to the length of $O P$, since $O P$ is perpendicular to $O R$ and $P Q$.

Therefore, the area of $O P Q R$ is $\frac{1}{2}(O R+P Q) \cdot O P=\frac{1}{2}(4+20) \cdot 8=96$.

(d) In general, quadrilateral $O P Q R$ is a trapezoid with parallel sides $O R$ and $P Q$, and height equal to the length of $O P$, since $O P$ is perpendicular to $O R$ and $P Q$.

In terms of $p$, the area of $O P Q R$ is

$$
\frac{1}{2}(O R+P Q) \cdot O P=\frac{1}{2}(4+(2 p+4)) \cdot p=\frac{1}{2}(2 p+8) p=p(p+4)
$$

If the area of $O P Q R$ is 77 , then $p(p+4)=77$ and so $p^{2}+4 p-77=0$ or $(p+11)(p-7)=0$.

Since $p>0$, then $p=7$.

We can verify that if $p=7$, then the area of $O P Q R$ is indeed 77 .