Thread: Equation of line HELP!

1. Equation of line HELP!

How do I work out.......

Find the equation of a straightline that makes an angle of 55 degrees with the negative X axis and cuts through (2,0)

and

Find the equation that runs perpendicular to 2x + 3y = 12 and cuts through the X axis at 2.

Thanks

2. Originally Posted by studiers
How do I work out.......

Find the equation of a straightline that makes an angle of 55 degrees with the negative X axis and cuts through (2,0)
You can determine a non-vertical line with a slope and point on it.

A point on it is (2,0) and the slope is m=tan55.

Use formula,
y-y_0=m(x-x_0)
y=tan55(x-2)
Simplify.

3. Hello, studiers!

Exactly where is your difficulty?

Do you know the "Point-Slope Formula"?

Given a point (x
1,y1) on the line and the slope of the line m,

. . the equation is: . y - y
1 .= .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