Hello, studiers!

Exactly where is your difficulty?

Do you know the "Point-Slope Formula"?

Given a point (x1,y1) on the line and the slope of the line *m*,

. . the equation is: . y - y1 .= .m(x - x1)

Assuming you're new to this material, I'll take baby-steps . . .

Find the equation of a straight line that makes an angle of 55°

with the negative x-axis and cuts through (2,0) Code:

* |
*
| *
| *
| *
| 55° * 125°
- - - + - - - - * - - -
| 2 *
| *

The line make an angle of θ = 125° with the positive x-axis.

Since slope is given by: .m = tanθ, we have: .m = tan125°

Point-Slope Formula: . y - 0 .= .tan125°(x - 2) . → . y .= .(tan125°)x - 2·tan125°

Find the equation that is perpendicular to 2x + 3y = 12

and cuts through the x-axis at 2. We are given: .2x + 3y .= .12 . → . y .= .(-2/3)x + 4

The line has slope: -2/3

The slope perpendicular to it is: .m = 3/2

The line through (2,0) with slope m = 3/2 is:

. . y - 0 .= .(3/2)(x - 2) . → . y .= .(3/2)x - 3