Question

The figure shows the graph of two perpendicular lines. Which of the following pairs of equations might have such a graph?

$$\begin{gathered} \left(a\right)y-2x=2 \\ y+2x=-1 \\ \left(b\right)y-2x=0 \\ 2y+x=0 \\ \left(c\right)2y-x=2 \\ 2y+x=-2 \\ \left(d\right)y-2x=2 \\ x+2y=-1 \\ \left(e\right)2x+y=-2 \\ 2y+x=-2 \end{gathered}$$

Short answer

Equation$$\begin{gathered} \left(d\right)y-2x=2 \\ x+2y=-1 \end{gathered}$$might have such graph

Step-by-step solution

Given information

We are given a graph and equation

Explanation for correct option

Consider the option $$\begin{gathered} \left(d\right)y-2x=2 \\ x+2y=-1 \end{gathered}$$

Simplifying the equation and writing it in slope intercept form we get

$$\begin{gathered} y=2x+2 \\ y=-\frac{1}{2}x-\frac{1}{2} \end{gathered}$$

On multiplying the slope we get $-1$

Hence the lines are perpendicular

Explanation for other options

Consider the equations

$$\begin{gathered} \left(a\right)y-2x=2 \\ y+2x=-1 \\ \left(b\right)y-2x=0 \\ 2y+x=0 \\ \left(c\right)2y-x=2 \\ 2y+x=-2 \\ \left(e\right)2x+y=-2 \\ 2y+x=-2 \end{gathered}$$

On simplifying the equation we get

$$\begin{gathered} \left(a\right)y=2x+2 \\ y=-2x-1 \\ \left(b\right)y=2x \\ y=-\frac{1}{2}x \\ \left(c\right)y=\frac{x}{2}+1 \\ y=-\frac{x}{2}-1 \\ \left(e\right)y=-2x-2 \\ y=-\frac{x}{2}-1 \end{gathered}$$

On multiplying the slopes of equation of lines in options a) ,c) ,d) we do not get $-1$

And for option b) on multiplying the slopes we do get $-1$but the lines passes through the origin hence the given graph cannot be described by the lines

Conclusion

The equation$$\begin{gathered} \left(d\right)y-2x=2 \\ x+2y=-1 \end{gathered}$$might have such graph.