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

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- Jan 31st 2009, 11:36 AMkeadyjrFinding the equation of a line
Find the slope-intercept equation of the line that passes through (-8, 9) and is perpendicular to

*x*- 8*y*= 2. - Jan 31st 2009, 11:44 AMred_dog
$\displaystyle x-8y=2\Rightarrow y=\frac{1}{8}x-\frac{1}{4}\Rightarrow m=\frac{1}{8}\Rightarrow m'=-\frac{1}{m}=-8$

Here m' is the slope of the perpendicular.

The equation is $\displaystyle y-y_0=m'(x-x_0)$ where $\displaystyle (x_0,y_0)=(-8,9)$ - Jan 31st 2009, 11:48 AMmasters
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:

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

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

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

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

$\displaystyle y=mx+b$

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

$\displaystyle b=-55$

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

$\displaystyle y=-8x-55$