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

1. An 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

2. 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 $\displaystyle m$, 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.

3. 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 $\displaystyle 9x+8y=3$ 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?

4. 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 line5, -2); 9x + 8y =3
Here is a useful trick.
Suppose that $\displaystyle A\ne 0~\&~B\ne 0$ then the lines
$\displaystyle Ax+By+C=0~\&~Bx-Ay+D=0$
are perpendicular.

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

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