Find the area of the triangle enclosed by the given equations: $y=-\frac12x-2, y=\frac13 x-2, $ and $ y=0$

Question

Find the area of the triangle enclosed by the given equations:

$y=-\frac12 x-2$
$y=\frac13 x-2$
$y=0$

Answer 1

The way to solve this is to find all the three points of intersection by solving two equations at a time simultaneously and then calculating the area of the triangle enclosed by the three points. Since y=0 is an equation, so you can put the value of y in the other two equations to easily get the value of x and get two points. To get the third point, solve the first two equations.

If the three points are, $$ (x_1,y_1),(x_2,y_2),(x_3,y_3) $$ , then area of the triangle enclosed is given by: $$ \frac {1} {2} \begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1 \end{vmatrix} $$

Answer 2

The intersection points are $(-4,0)$, $(6,0)$ and $(0,-2)$. The distance between the points on the line $y=0$ is $10$; the distance from $(0,-2)$ to the line $y=0$ is $2$.

Can you finish?