Find the intersection of 2 lines
$$y=2x-2\ and\ y=-\frac{x}{2}+8$$
$$2x-2=-\frac{x}{2}+8$$
$$4x-4=-x+16$$
$$5x=20$$ $$x=4$$
solve for y in y=2x-2
$$y=2\left(4\right)-2$$ , $$y=8-2=6$$ (height of the triangle)
Since y=0 because we are dealing with x-axis will be
$$y=2x-2$$
$$0=2x-2$$
$$x=1$$
solve for $$x\ \ in\ y=-\frac{x}{2}+8$$
$$0=-\frac{x}{2}+8$$
$$\frac{x}{2}=8$$
$$x=16$$
$$Area\ of\ Triangle=\frac{1}{\frac{2}{ }}\cdot b\cdot h$$
$$Area\ of\ Triangle=\frac{1}{\frac{2}{ }}\cdot15\cdot6=45\ $$
Answer is $$Option\ D$$

swerve wrote:What is the area of the triangle formed by the intersection of lines y=2x-2, y=-x/2 +8 and y=0?

A. 20
B. 30
C. 40
D. 45
E. 60
Source: Veritas Prep

\[? = {S_{\Delta {\text{ABC}}}}\]

\[{\text{point}}\,\,C\,\,\,\left\{ \begin{gathered}
\,y = 2x - 2\,\,\,\, \hfill \\
\,y = - \frac{x}{2} + 8\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\,4} \,\,\,\,\,4y = - 2x + 32 \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\,\,\,\,\,5y = 30\,\,\,\,\, \Rightarrow \,\,\,\,\,y = 6\,\,\,\,\,\,\,\left( { \Rightarrow \,\,\,\,\,x = 4} \right)\]
\[? = \frac{{AB \cdot CD}}{2} = \frac{{\left( {16 - 1} \right) \cdot 6}}{2} = 45\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.