An equation of the line containing the given point and perpendicular to the line

**jay1** · May 12th 2011, 09:28 AM

I need help with the steps and solving for an equation of the line containing the given point and perpendicular to the line:
(5, -2); 9x + 8y =3
I can find the slope (-8/9x?)... I just can't figure out the rest

**Quacky** · May 12th 2011, 09:33 AM

Originally Posted by jay1

I need help with the steps and solving for an equation of the line containing the given point and perpendicular to the line:
(5, -2); 9x + 8y =3
I can find the slope (-8/9x?)... I just can't figure out the rest

The gradient of the normal to the line will be the negative reciprocal of the original gradient. That is, if your gradient is $m$ , the gradient of the normal will be $\frac{-1}{m}$ . Then, once you've found the gradient, substitute it, along with your point, into $y-y_1=m(x-x_1)$ or $y=mx+c$ to find the equation.

**masters** · May 12th 2011, 09:48 AM

Originally Posted by jay1

I need help with the steps and solving for an equation of the line containing the given point and perpendicular to the line:
(5, -2); 9x + 8y =3
I can find the slope (-8/9x?)... I just can't figure out the rest

Hi jay1,

The slope of $9x+8y=3$ is not $-\frac{8}{9}x$

Solve for y to put the equation in slope-intercept form ( $y=mx+b$ where $m$ is the slope):

$8y=-9x+3$

$y=-\frac{9}{8}x+\frac{3}{8}$

Now do you know what the slope is?

**Plato** · May 12th 2011, 11:18 AM

Originally Posted by jay1

I need help with the steps and solving for an equation of the line containing the given point and perpendicular to the line

5, -2); 9x + 8y =3

Here is a useful trick.
Suppose that $A\ne 0~\&~B\ne 0$ then the lines
$Ax+By+C=0~\&~Bx-Ay+D=0$
are perpendicular.

So for your problem just use your point in $8x-9y=k$ to find the value of k.

**Wilmer** · May 12th 2011, 12:18 PM

You've got a point and the slope; go find out:
Straight-Line Equations: Point-Slope Form

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