SOLUTION: Write the standard form of the line that passes through (5,-3) and has a slope perpendicular to 2x+7y=16.

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Question 120206: Write the standard form of the line that passes through (5,-3) and has a slope perpendicular to 2x+7y=16.
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
First step is to find the required slope value. Perpendicular lines have slopes that are negative reciprocals of each other: m%5B1%5D=-%281%2Fm%5B2%5D%29

The equation of the perpendicular is 2x%2B7y=16. Put this into slope-intercept form, y=mx%2Bb, by solving for y.

2x%2B7y=16
7y=-2x%2B16
y=%28%28-2%29%2F7%29x+%2B16, which tells us that the slope of the perpendicular to the desired line is -%282%2F7%29.

Using m%5B1%5D=-%281%2Fm%5B2%5D%29, we now know that the slope of the desired line is 7%2F2.

Now we have a point, (5,-3) and a slope, 7%2F2, so use the point slope form of the line:

%28y-y%5B1%5D%29=m%28x-x%5B1%5D%29 where m is the slope and (x%5B1%5D,y%5B1%5D) are the coordinates of the given point.

y-%28-3%29=%287%2F2%29%28x-5%29%29

Now simplify to standard form:

Multiply by 2

2y%2B6=7%28x-5%29

Distribute:

2y%2B6=7x-35

Add -7x to both sides

-7x%2B2y%2B6=-35

Add -6 to both sides:

-7x%2B2y=-41 is the desired equation, though it would be a little tidier to multiply through by -1:

7x-2y=41


Hope this helps,
John