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Question 1182992: Find the equation of the straight line which
passes through the point(-3,5) and is
perpendicular to the line 2x-4y+3=0?
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Using basic algebra....
(1) Find the slope of the given line
2x-4y+3=0
4y = 2x+3
y = (1/2)x+3/4
The slope is 1/2; the slope of a line perpendicular to that line is -2/1, or just -2.
(2) Use the general slope-intercept form of the equation with the desired slope of -2 passing through (x,y) = (-3,5) to find the equation
y = mx+b
5 = (-2)(-3)+b
5 = 6+b
b = -1
ANSWER: y = -2x-1
A faster path to the answer....
Given the equation 2x-4y+3=0, any line parallel to that line has the form 2x-4y+C=0 for some constant C, and any line perpendicular to that line has the form 4x+2y+C=0 for some constant C (switch the coefficients of x and y and change the sign of one of them).
So we are looking for an equation of the form 4x+2y+C=0 that is satisfied by (x,y) = (-3,5):
4(-3)+2(5)+C=0
-12+10+C=0
C=2
The equation we want is
4x+2y+2=0
That is equivalent to the answer we got using basic algebra:
4x+2y+2=0
2y = -4x-2
y = -2x-1
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