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Q. 132

Expert-verifiedFound in: Page 44

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

We are given a graph and equation

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

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

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

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