Where does Line L intersect with Line M?

Question

Line L is perpendicular to line 6x-y=7 and passes through the point (0,6). Line M is parallel to the line y=2/3x-4 and passes through the point (-3,-1). Where do these lines intersect? Please explain your answer.

Answer 1

6x - y = 7 can be written as: 6x - 7 = y, so its slope is 6. The perpendicular slope for L would therefore be -1/6. Using the point-slope equation:

slope = -1/6, point = (0, 6)

-1/6 = (y - 6)/(x - 0)

Cross-multiply:

-x = 6(y - 6) = 6y - 36

Add x and 37 to both sides:

36 = x + 6y

Line M is parallel to y = 2/3 x - 4, so it must have the same slope, 2/3.

slope = 2/3, point = (-3, -1)

2/3 = (y + 1)/(x + 3)

Cross multiply:

2(x + 3) = 3(y + 1)

Distribute:

2x + 6 = 3y + 3

Subtract 3y and 6 from both sides:

2x - 3y = -3

So now the two equations are:

x + 6y = 36
2x - 3y = -3

Multiply the second equation by 2 to make it easier to solve by addition:

x + 6y = 36
4x - 6y = -6
-----------------
x + 4x + 6y - 6y = 36 - 6

Simplify:

5x = 30

Divide both sides by 5 to get x:

x = 6

Plug x back into one of the other equations to get y:

2(6) - 3y = -3
12 - 3y = -3
-3y = -15
y = 5

So the point would be (6, 5).

Double check:

(0, 6):
0 + 6(6) = 36, check!

(-3, -1):
2(-3) - 3(-1) = -6 - (-3) = -3, check!

(6, 5):
6 + 6(5) = 6 + 30 = 36, check!
2(6) - 3(5) = 12 - 15 = -3, check!