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
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 $\displaystyle m$, 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.
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?
You've got a point and the slope; go find out:
Straight-Line Equations: Point-Slope Form