Consider the lines with equations $y=m x$ and $y=-2 x+28$ where $m$ is some real number. The area enclosed by the two lines and the $x$-axis in the first quadrant is equal to 98 . What is the value of $m$ ?
The area of interest is a triangle with vertices at the origin, the $x$-intercept of $y=-2 x+28$, and the point of intersection of the lines with equations $y=m x$ and $y=-2 x+28$.

![](https://cdn.mathpix.com/cropped/2024_04_17_54c374478de6843636d1g-08.jpg?height=474&width=542&top_left_y=1595&top_left_x=843)

Setting $y=0$ in $y=-2 x+28$ gives $0=-2 x+28$ which implies $x=14$.

Therefore, the base of the triangle has length $14-0=14$.

Setting $m x=-2 x+28$ leads to $x=\frac{28}{m+2}$, and substituting this into $y=m x$ gives $y=\frac{28 m}{m+2}$, which is the height of the triangle.

Using that the area of the triangle is 98 , we have that $98=\frac{1}{2} \times 14 \times \frac{28 m}{m+2}$ or $98=\frac{196 m}{m+2}$.

Dividing both sides of this equation by 98 gives $1=\frac{2 m}{m+2}$, which can be rearranged to get $m+2=2 m$, which implies that $m=2$.