Question

each side of a triangle lies along a line in a coordinate plane. the three lines that contain the sides are represented by the equations given to the right. (assume ( x - 2y=-4), ( 2x + y=-3), ( 7x - 4y=12))

b. what are the coordinates of the vertices?
the coordinates of the vertices are (square) (use a comma to separate answers as needed. type an ordered pair.)

Explanation:

Step1: Solve the system

$$\begin{cases}x - 2y=-4\\2x + y=-3\end{cases}$$

From the first equation $x=2y - 4$. Substitute $x = 2y-4$ into the second equation:
$$2(2y - 4)+y=-3$$
$$4y-8 + y=-3$$
$$5y=5$$
$$y = 1$$
Substitute $y = 1$ into $x=2y - 4$, we get $x=2\times1-4=-2$.

Step2: Solve the system

$$\begin{cases}x - 2y=-4\\7x-4y=-3\end{cases}$$

Multiply the first equation by $2$: $2x-4y=-8$.
Subtract $2x - 4y=-8$ from $7x-4y=-3$:
$$(7x-4y)-(2x - 4y)=-3-(-8)$$
$$7x-4y - 2x + 4y=5$$
$$5x=5$$
$$x = 1$$
Substitute $x = 1$ into $x-2y=-4$, we get $1-2y=-4$, $-2y=-5$, $y=\frac{5}{2}$.

Step3: Solve the system

$$\begin{cases}2x + y=-3\\7x-4y=-3\end{cases}$$

Multiply the first equation by $4$: $8x+4y=-12$.
Add $8x + 4y=-12$ to $7x-4y=-3$:
$$(8x+4y)+(7x-4y)=-12+(-3)$$
$$15x=-15$$
$$x=-1$$
Substitute $x=-1$ into $2x + y=-3$, we get $-2 + y=-3$, $y=-1$.

Answer:

$(-2,1),(1,\frac{5}{2}),(-1,-1)$