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Q. 132

Expert-verified
Found in: Page 44

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# The figure shows the graph of two perpendicular lines. Which of the following pairs of equations might have such a graph?$\left(a\right)y-2x=2\phantom{\rule{0ex}{0ex}}y+2x=-1\phantom{\rule{0ex}{0ex}}\left(b\right)y-2x=0\phantom{\rule{0ex}{0ex}}2y+x=0\phantom{\rule{0ex}{0ex}}\left(c\right)2y-x=2\phantom{\rule{0ex}{0ex}}2y+x=-2\phantom{\rule{0ex}{0ex}}\left(d\right)y-2x=2\phantom{\rule{0ex}{0ex}}x+2y=-1\phantom{\rule{0ex}{0ex}}\left(e\right)2x+y=-2\phantom{\rule{0ex}{0ex}}2y+x=-2$

Equation $\left(d\right)y-2x=2\phantom{\rule{0ex}{0ex}}x+2y=-1$might have such graph

See the step by step solution

## Step 1: Given information

We are given a graph and equation

## Step 2: Explanation for correct option

Consider the option $\left(d\right)y-2x=2\phantom{\rule{0ex}{0ex}}x+2y=-1$

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

$y=2x+2\phantom{\rule{0ex}{0ex}}y=-\frac{1}{2}x-\frac{1}{2}$

On multiplying the slope we get $-1$

Hence the lines are perpendicular

## Step 3: Explanation for other options

Consider the equations

$\left(a\right)y-2x=2\phantom{\rule{0ex}{0ex}}y+2x=-1\phantom{\rule{0ex}{0ex}}\left(b\right)y-2x=0\phantom{\rule{0ex}{0ex}}2y+x=0\phantom{\rule{0ex}{0ex}}\left(c\right)2y-x=2\phantom{\rule{0ex}{0ex}}2y+x=-2\phantom{\rule{0ex}{0ex}}\left(e\right)2x+y=-2\phantom{\rule{0ex}{0ex}}2y+x=-2$

On simplifying the equation we get

$\left(a\right)y=2x+2\phantom{\rule{0ex}{0ex}}y=-2x-1\phantom{\rule{0ex}{0ex}}\left(b\right)y=2x\phantom{\rule{0ex}{0ex}}y=-\frac{1}{2}x\phantom{\rule{0ex}{0ex}}\left(c\right)y=\frac{x}{2}+1\phantom{\rule{0ex}{0ex}}y=-\frac{x}{2}-1\phantom{\rule{0ex}{0ex}}\left(e\right)y=-2x-2\phantom{\rule{0ex}{0ex}}y=-\frac{x}{2}-1$

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 $\left(d\right)y-2x=2\phantom{\rule{0ex}{0ex}}x+2y=-1$might have such graph.