1. parallel to 2x – 3y + 11 = 0 going through (2,–1)
2x - 3y + 11 = 0
(2/3)x + 11/3 = y
Slope = 2/3 Parallel line has the same slope
Now use point slope form to determine the equation of that line
y-2 = (2/3)(x+1)
y = (2/3)x + 2/3 + 2
y = (2/3)x +8/3
2. parallel to 3x + y = 2y – 4x + 7 going through (–2,–7)
3x + y = 2y - 4x + 7
7x - 7 = y
Slope = 7
y+2 = 7(x+7)
y = 7x + 49 - 2
y = 7x + 47
1. Perpendicular to 3y – 2x + 11 = 0 going through (2,–2)
From 1) above the slope of this line is 2/3. So, the slope of a line perpendicular to this line is the negative inverse or -3/2
y+2 = (-3/2)(x-2)
y = (-3/2)x + 3 - 2
y = (-3/2)x + 1
2. Find k if the lines ""kx – 2y + 5 = 0 and 3x + 4y + 1 = 0"" are:
a. perpendicular b. parallel
a. perpendicular b. parallel
3x + 4y + 1 = 0
3x+ 1 = -4y
(-3/4)x -1/4 = y
So the slope of this line is -3/4
a) And the slope of the perpendicular line (kx - 2y + 5 = 0) is 4/3
Let's work this through.
kx + 5 = 2y
(k/2)x + 5/2 = y
k/2 = 4/3
k = 8/3
b) For parallel lines, the slope is -3/4
k/2 = -3/4
k = -3/2
Landen B.
What is the answer to 3x+2y=8 and 2x+3y=-12
Report
05/12/21
Michael J.
01/10/16