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What is the equation of a line that is perpendicular to x+2y=4 and passes through the point (−4, -7) ?

User Tanz
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Answer
So, the equation of the line perpendicular to x+2y=4 and passing through the point (−4,−7) is y=2x+1.
Step by Step

To find the equation of a line perpendicular to x+2y=4 and passing through the point (−4,−7), we'll first find the slope of the given line and then determine the negative reciprocal to get the slope of the perpendicular line.

The Equation x+2y=4 can be written in slope-intercept form (y=mx+b) by solving for y:

x+2y=4

2y=−x+4

y=-1/2x+2

So, the slope of the given line is -1/2 The negative reciprocal of -1/2 is 2 which will be the slope of the perpendicular line.

Now, we can use the point-slope form of a line to find the equation of the perpendicular line:

y-y1=m(x-x1)

Using the point (−4,−7) and the slope m=2

y−(−7)=2(x−(−4))

Simplify the equation: y+7=2(x+4)

Distribute 2 on the right side: y+7=2x+8

Subtract 7 from both sides to isolate y=2x+1

So, the equation of the line perpendicular to x+2y=4 and passing through the point (−4,−7) is y=2x+1.

User Adam Mitz
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