Finding the equation of a line

**keadyjr** · January 31st 2009, 11:36 AM

Find the slope-intercept equation of the line that passes through (-8, 9) and is perpendicular to x - 8y = 2.

**red_dog** · January 31st 2009, 11:44 AM

$x-8y=2\Rightarrow y=\frac{1}{8}x-\frac{1}{4}\Rightarrow m=\frac{1}{8}\Rightarrow m'=-\frac{1}{m}=-8$

Here m' is the slope of the perpendicular.
The equation is $y-y_0=m'(x-x_0)$ where $(x_0,y_0)=(-8,9)$

**masters** · January 31st 2009, 11:48 AM

Originally Posted by keadyjr

Find the slope-intercept equation of the line that passes through (-8, 9) and is perpendicular to x - 8y = 2.

Hi keadyjr,

First of all find the slope of your line given by the equation x - 8y = 2. Let's put it in slope-intercept form:

$y=\frac{1}{8}x-\frac{1}{4}$

So the slope is $\frac{1}{8}$

The slope of a line perpendicular to this line would be the negative reciprocal of $\frac{1}{8}$ , or $-8$ .

Using the point (-8, 9) and a slope of -8, we can now find the y-intercept.

$y=mx+b$

$9=-8(-8)+b$

$b=-55$

Substituting back into the slope intercept form, we arrive at:

$y=-8x-55$

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