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(Headbang)

- May 12th 2011, 09:28 AMjay1An equation of the line containing the given point and perpendicular to the line
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(Headbang) - May 12th 2011, 09:33 AMQuacky
The gradient of the normal to the line will be the negative reciprocal of the original gradient. That is, if your gradient is $\displaystyle m$, the gradient of the normal will be $\displaystyle \frac{-1}{m}$. Then, once you've found the gradient, substitute it, along with your point, into $\displaystyle y-y_1=m(x-x_1)$ or $\displaystyle y=mx+c$ to find the equation.

- May 12th 2011, 09:48 AMmasters
Hi jay1,

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

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

$\displaystyle 8y=-9x+3$

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

Now do you know what the slope is?

- May 12th 2011, 11:18 AMPlato
- May 12th 2011, 12:18 PMWilmer
You've got a point and the slope; go find out:

Straight-Line Equations: Point-Slope Form