Asked by Jocelyne
Two lines L1 and L2, are perpendicular. The equation of L1 is 3x-y=2. L2 passes through (-5, -1)
Answers
Answered by
Helper
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
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
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