2. The line $L_{1}$ has equation $y=-\frac{4}{3} x$ and passes through the origin, $O$. The line $L_{2}$ has equation $y=-\frac{1}{2} x+5$ and crosses the $x$-axis at $P$. Lines $L_{1}$ and $L_{2}$ intersect at $Q$.
(a) What are the coordinates of points $P$ and $Q$ ? (No justification is required.)
(b) Find the area of $\triangle O P Q$.
(c) Point $R$ is on the positive $x$-axis so that the area of $\triangle O Q R$ is three times the area of $\triangle O P Q$. Determine the coordinates of $R$.
(d) Point $S$ has coordinates $(18, t)$, with $t>0$. If the area of $\triangle O Q S$ is three times the area of $\triangle O P Q$, determine the value of $t$.
2. (a) To find the coordinates of $P$, the point where line $L_{2}$ crosses the $x$-axis, we set $y=0$ and solve for $x$.
Thus, we solve $0=-\frac{1}{2} x+5$ or $\frac{1}{2} x=5$ to get $x=10$.
Therefore, the coordinates of $P$ are $(10,0)$.

To find the coordinates of $Q$, the point of intersection of lines $L_{1}$ and $L_{2}$, we equate values of $y$ and solve for $x$.
Thus, we solve $-\frac{4}{3} x=-\frac{1}{2} x+5$. We multiply both sides by 6 to obtain $6\left(-\frac{4}{3} x\right)=6\left(-\frac{1}{2} x+5\right)$ or $-8 x=-3 x+30$, which gives $-5 x=30$ or $x=-6$.
To find the $y$-coordinate of $Q$, we substitute $x=-6$ into the equation of line $L_{1}$ to obtain $y=-\frac{4}{3}(-6)=8$.
Therefore, the coordinates of $Q$ are $(-6,8)$.
(b) The coordinates of the vertices of $\triangle O P Q$ are $O(0,0), P(10,0)$, and $Q(-6,8)$.
We can view this triangle as having base $O P$ along the $x$-axis. $O P$ has length 10 .
The height of this triangle is then the distance from $Q$ to the $x$-axis, or 8 .
Therefore, the area $\triangle O P Q$ is $\frac{1}{2}(10)(8)=40$.
(c) Since the area of $\triangle O P Q$ is 40 and the area of $\triangle O Q R$ is three times the area of $\triangle O P Q$, then the area of $\triangle O Q R$ is $3 \times 40=120$.
Since $R$ is to be on the positive $x$-axis, then we can view $\triangle O Q R$ as having base $O R$ along the $x$-axis. Suppose that the length of the base is $b$.
The height of this triangle is again the distance from $Q$ to the $x$-axis, or 8 .
Thus, we want $\frac{1}{2} b(8)=120$ or $4 b=120$, and so $b=30$.
Since $O$ has coordinates $(0,0)$ and $R$ is on the positive $x$-axis, then $R$ must have coordinates $(30,0)$.
(We could also have noted that $\triangle O Q R$ and $\triangle O P Q$ can be viewed as having the same height, so for the areas to be in the ratio $3: 1$, then the ratio of the lengths of the bases must also be $3: 1$.)
(d) Since the area of $\triangle O Q S$ is to be three times the area of $\triangle O P Q$, then its area is to be 120 , which is equal to the area of $\triangle O Q R$ from (c). Here are two methods to find the value of $t$.
Method 1: Constant heights
Consider $\triangle O Q S$ and $\triangle O Q R$.
These triangles are to have the same area, and can be viewed as having a common base, $O Q$.
Therefore, the two triangles must have the same height.
This height will be the perpendicular distance from $R(30,0)$ to the line $y=-\frac{4}{3} x$, along which $O Q$ lies.
We find the coordinates of $S$ (knowing that it must have an $x$-coordinate of 18) by sliding a point from $R$ along a line parallel to the base $O Q$ (so along a line with slope $-\frac{4}{3}$ ). This process will maintain the height.

To get from $R$ to $S$, we slide $30-18=12$ units to the left. Since the slope of the line is $-\frac{4}{3}$, then we must go up by $\frac{4}{3}(12)=16$ units.
Therefore, $S$ has coordinates $(18,16)$; that is, $t=16$.
Method 2: Using a trapezoid
Join points $O, Q$ and $S$ to each other, and draw perpendiculars from $Q$ to $A(-6,0)$ and from $S$ to $B(18,0)$ on the $x$-axis.
Then the area of trapezoid $A Q S B$ will equal the sum of the areas of $\triangle A Q O, \triangle O Q S$ and $\triangle O S B$.
$\triangle A Q O$ has base 6 and height 8 (since the coordinates of $Q$ are $(-6,8)$ ) and so $\triangle A Q O$ has area $\frac{1}{2}(6)(8)=24$.
$\triangle O Q S$ is to have area 120 .
$\triangle O S B$ has base 18 and height $t$ (since the coordinates of $S$ are $(18, t))$ and so $\triangle O S B$ has area $\frac{1}{2}(18) t=9 t$.
Thus, trapezoid $A Q S B$ has area $24+120+9 t$.
Viewed another way, trapezoid $A Q S B$ has parallel sides $A Q$ and $B S$ of lengths 8 and $t$, and height $A B$ of length $6+18=24$, and so has area $\frac{1}{2}(8+t)(24)=96+12 t$.
Equating these areas, we obtain $96+12 t=24+120+9 t$ and so $3 t=48$ or $t=16$.