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May 5, 2016

# Homework Help: Algebra 1

Posted by Jocelyne on Monday, February 21, 2011 at 1:03pm.

Two lines L1 and L2, are perpendicular. The equation of L1 is 3x-y=2. L2 passes through (-5, -1)
• Algebra 1 - Helper, Monday, February 21, 2011 at 4:10pm

Do you need to find L2?

If two lines are perpendicular, the slope m of one is the negative reciprocal of the slope of the other,

Slope m1 * m2 = -1
or, slope m2 = -1/m1

Your need to find slope m1 to write the equation of the perpendicular line.

L1 = 3x - y = 2
To find slope m, put the equation in slope-intercept form
y = mx + b, where m = slope and b = y-intercept

3x - y = 2
y = 3x - 2
So, slope m1 = 3

Since, m2 = -1/m1 and m1 = 3,
m2 = -1/3 = slope of L2 (perpendicular line)

L2 through point (-5, -1)
Form of the equation is,
y = mx + b

You found the slope m2 = -1/3
y = -1/3 x + b

To find b, use point (-5, -1) and substitute x and y point values in the equation and solve for b
y = -1/3 x + b
y = -5, x = -1
-1 = -1/3 (-5) + b
-1 = 5/3 + b
b = -1 + -5/3
b = -3/3 + -5/3
b = -8/3

So, L2 is
y = -1/3 x + -8/3
y = -1/3 x - 8/3