What is the slope of a line perpendicular to #x+2y=7#?

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Answer 1 · 2018-07-06T23:10:38

The slope perpendicular to the equation is 2.

First, find the slope of the equation by setting it into slope-intercept form, #y = mx + b#:
#x + 2y = 7#

Subtract #color(blue)x# from both sides:
#x + 2y quadcolor(blue)(-quadx) = 7 quadcolor(blue)(-quadx)#

#2y = 7 - x#

Divide both sides by #color(blue)2#:
#(2y)/color(blue)2 = (7-x)/color(blue)2#

#y = 7/2-1/2x#

We know that the slope of a slope-intercept equation is the value multiplied by #x#. That means the slope of the equation is #-1/2#.

Now to find the slope perpendicular to the equation, we find the negative reciprocal, or switching the sign and doing #1# over the slope.

So:
#(-1)/(-1/2) = -2#

Therefore, the slope perpendicular to the equation is 2.

Hope this helps!