SOLUTION: Write the equation of a line that goes through the point (4, -3) and is perpendicular to the line x−4y=16

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Question 1157269: Write the equation of a line that goes through the point (4, -3) and is perpendicular to the line x−4y=16
Found 2 solutions by Boreal, MathLover1:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
x-4y=16 can be written as 4y+16=x or 4y=x-16 or y=(1/4)x-4
the slope of that line is 1/4
perpendicular lines have negative reciprocal slopes when multiplied together, so the line wanted has slope -4.
it goes through (4, -3)
point slope formula y-y1=m(x-x1), m slope (x1, y1) point
y+3=-4(x-4)
y=-4x+13
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C%28x%2F4%29-4%2C-4x%2B13%29

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

the equation of a line that goes through the point (4, -3) and is perpendicular to the line x-4y=16
perpendicular lines have slopes negative reciprocal to each other
so, first find a slope of given line:
x-4y=16.......solve for y
x-16=4y
x%2F4-16%2F4=y
y=%281%2F4%29x-4=> a slope is 1%2F4
then negative reciprocal is -1%2F%281%2F4%29=-4
=>perpendicular line have slope m=-4

so far, the equation of a line will be

y=-4x%2Bb...........use iven point (4, -3) to calculate b

-3=-4%2A4%2Bb
-3=-16%2Bb
-3%2B16=b
13=b

and, your equation is:
y=-4x%2B13