Algebra 1
posted by Jocelyne .
Two lines L1 and L2, are perpendicular. The equation of L1 is 3xy=2. L2 passes through (5, 1)

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 slopeintercept form
y = mx + b, where m = slope and b = yintercept
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