need help with stumper

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• Sep 18th 2005, 04:07 PM
benben
need help with stumper
How would I write an equation of the line that passes through (-2, 3) and is perpendicular to the line y= 2x +1
• Sep 18th 2005, 10:23 PM
rgep
Let's write lines in the form y = mx+c. If your line passes through the point with x=-2, y=3 then 3 = -2m + c. Now if lines y = mx + c and y = nx + d are perpdendicular then the slopes m and n are related by mn = -1. So your line has slope m = -1/2, being perpendicular to a line with slope 2. That gives you c=2.
• Sep 18th 2005, 10:43 PM
ticbol
Here is one way.

You have one point on the unknown line. So if you can find the slope of this line, you can then find this unknown line by using the point-slope form of the equation of a line.
(y -y1) = m(x -x1) ------***
where (x1,y1) is the pnown point.

The unknown line is perpendicular to the known line.
You know that when two lines are perpendicular, their slopes are negative reciprocals.
m2 = -1/m1, or m1 = -1/m2 -------***

In the slope-intercept form of the equation of a line, y = mx +b, the slope of the line is the "m".
So in the given line,
y = 2x +1
its slope is the "2"
So m1 = 2
Therefore the slope of the unknown line is m2 = -1/m1 = -1/2.

Then , the unknown line is
(y -3) = (-1/2)(x -(-2))
y -3 = -(1/2)(x +2)
y -3 = -(1/2)x -1
y = -(1/2)x -1 +3
y = -(1/2)x +2 -----answer, in y=mx+b form.

If you want it in general form Ax +Bx +C = 0,
y = -(1/2)x +2
Clear the fraction, multiple both sides by 2,
2y = -x +4
x +2y -4 = 0 ------answer also.