The line $y=-\frac{3}{4} x+9$ crosses the $x$-axis at $P$ and the $y$-axis at $Q$. Point $T(r, s)$ is on line segment $P Q$. If the area of $\triangle P O Q$ is three times the area of $\triangle T O P$, then the value of $r+s$ is
(A) 7
(B) 10
(C) 11
(D) 14
(E) 18

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-106.jpg?height=466&width=447&top_left_y=564&top_left_x=1294)

Part C: Each correct answer is worth 8.
Solution 1

The $y$-intercept of the line $y=-\frac{3}{4} x+9$ is $y=9$, so $Q$ has coordinates $(0,9)$.

To determine the $x$-intercept, we set $y=0$, and so obtain $0=-\frac{3}{4} x+9$ or $\frac{3}{4} x=9$ or $x=12$. Thus, $P$ has coordinates $(12,0)$.

Therefore, the area of $\triangle P O Q$ is $\frac{1}{2}(12)(9)=54$, since $\triangle P O Q$ is right-angled at $O$.

Since we would like the area of $\triangle T O P$ to be $\frac{1}{3}$ that of $\triangle P O Q$, then the area of $\triangle T O P$ should be 18 .

If $T$ has coordinates $(r, s)$, then $\triangle T O P$ has base $T O$ of length 12 and height $s$, so $\frac{1}{2}(12)(s)=18$ or $6 s=18$ or $s=3$.

Since $T$ lies on the line, then $s=-\frac{3}{4} r+9$ or $3=-\frac{3}{4} r+9$ or $\frac{3}{4} r=6$ or $r=8$.

Thus, $r+s=8+3=11$.

## Solution 2

The $y$-intercept of the line $y=-\frac{3}{4} x+9$ is $y=9$, so $Q$ has coordinates $(0,9)$.

To determine the $x$-intercept, we set $y=0$, and so obtain $0=-\frac{3}{4} x+9$ or $\frac{3}{4} x=9$ or $x=12$. Thus, $P$ has coordinates $(12,0)$.

The areas of triangles are proportional to their heights if the bases are equal.

Since the area of $\triangle P O Q$ is 3 times the area of $\triangle T O P$, then the height of $\triangle T O P$ is $\frac{1}{3}$ that of $\triangle P O Q$.

Thus, $T$ is $\frac{1}{3}$ along $P Q$ from $P$ towards $Q$.

Since $P$ has coordinates $(12,0)$ and $Q$ has coordinates $(0,9)$, then $T$ has coordinates $\left(\frac{2}{3}(12), \frac{1}{3}(9)\right)=(8,3)$.

Thus, $r+s=8+3=11$.

ANswer: (C)

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