Equation of a line (point/perpendicular line)

You can view as many worked out examples as you want. First you are shown the question. Click Show Answer to view the answer.
Click Show another Example to view another example. The best way to master mathematics is by practice. But practice requires time.
If you don't have the time to practice, you can always view many practice problems completely worked out and become good at it too.

Find the equation of the line passsing through the point P $\left(\frac{-8}{3},3\right)$ and parallel to the line $5x+4y+6= 0$

**Answer**

The slope of the line $5x+4y+6= 0$ is given by $m = -\frac{5}{4}$

The second line is parallel to this line. So, they have the same slopes $m=-\frac{5}{4}$

The second line passes through the point $\left(\frac{-8}{3},3\right)$ and has slope $m=-\frac{5}{4}$

The equation of the line with slope $m$ and passing through the point $\left(x_0,y_0\right)$ is given by $$y-y_0=m\left(x-x_0\right)$$

Here $m=-\frac{5}{4}$, $x_0=-\frac{8}{3}$ and $y_0=3$

Substituting these values in the equation, we get

$ y-\left(3\right)=-\frac{5}{4}\left(x-\left(-\frac{8}{3}\right)\right)$

or, $ y-3=-\frac{5}{4}\left(x+\frac{8}{3}\right)$

or, $y-3=-\frac{5}{4}x-\frac{10}{3}$

or, $y=-\frac{5}{4}x-\frac{1}{3}$