Question

What system of equations is shown on the graph below?


On a coordinate plane, a line goes through points (negative 2, 0) and (0, negative 4) and another line goes through (0, negative 2) and (4, 0).

x minus 2 y = negative 4 and 2 x + y = 4
x minus 2 y = 4 and 2 x + y = negative 4
x minus 2 y = negative 2 and 2 x + y = negative 4
x minus 2 y = 2 and 2 x + y = 4

Answer 1

Answer:

The equation of the line passing through the points (-2,0) and (0,-4)

$\frac{y - y1}{x - x1} = \frac{y2 - y1}{x2 - x1}$

$\frac{y - 0}{x - ( - 2)} = \frac{ - 4 - 0}{0 - ( - 2)} = \frac{ - 4}{2} = - 2$

$y = - 2(x + 2) \\ y = - 2x - 4$

$2x + y = - 4$

The equation of the line passing through the points (0,-2) and (4,0)

$\frac{y - y1}{x - x1} = \frac{y2 - y1}{x2 - x1} \:$

$\frac{y - ( - 2)}{x - 0} = \frac{0 - ( - 2)}{4 - 0}$

$\frac{y + 2}{x} = \frac{1}{2}$

$x - 2y = 4$

Option B

x minus 2 y = 4 and 2 x + y = negative 4

Answer 2

The system of equation can be determine by using formula of equation of the line passing through the points $(x_1, y_1)$ and $(x_2, y_2)$.

The system equation would be $2x+y=-4$ and $x-2y=4$.

Given:

The given points are $(-2,0)$ and $(0,-4)$ and for another line $(0,-2)$ and $(4,0)$.

Find the equation line passing through point $(-2,0)$ and $(0,-4)$ .

$\dfrac{y-y_1}{x-x_1}=\dfrac{y_2-y_1}{x_2-x_1}$

Substitute the value in above equation.

$\dfrac{y-0}{x-(-2)}=\dfrac{-4-0}{0-(-2)}\\y=-2(x+2)\\y=-x-4\\2x+y=-4$

Find the equation line passing through point $(0,-2)$ and $(4,0)$ .

$\dfrac{y-y_1}{x-x_1}=\dfrac{y_2-y_1}{x_2-x_1}$

Substitute the value in above equation.

$\dfrac{y-(-2)}{x-0}=\dfrac{0-(-2)}{4-0}\\\\x-2y=4$

Thus, the system equation would be $2x+y=-4$ and $x-2y=4$.

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