Q. 132 - Precalculus Enhanced with Graphing Utilities

Question

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

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

Step-by-Step Solution

Answer

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

Step 1: Given information

We are given a graph and equation

Step 2: 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

Step 3: 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

Step 4: Conclusion

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