Math Problem Solver

Standard Form:

Perpendicular lines are two lines in the same plane which intersect at a 90 degree angle. Perpendicular lines have slopes that are negative reciprocals of each other, so the slope of the given line can be used to find the slope of a line perpendicular to it.$$\begin{aligned}Ax + By + C &= 0 \\ 3x -2y +28 &= 0 \end{aligned}$$$$\begin{aligned} \text{Slope} &= -\frac{A}{B} \\ \text{Slope} &= \frac{3}{2} \end{aligned}$$The negative reciprocal of $\frac{3}{2}$ is $- \frac{2}{3}$, therefore $A$ and $B$ of a perpendicular line are $-2$ and $-3$.$$\begin{aligned} -2x -3y + C &= 0 \\ -2(-5) -3(20)+C &= 0 \\ 10 + C &= 60 \\ C &= 50 \end{aligned}$$An equation of the line in standard form is: $-2x -3y +50 = 0$.

Slope-Intercept Form:

Perpendicular lines are two lines in the same plane which intersect at a 90 degree angle. Perpendicular lines have slopes that are negative reciprocals of each other, so the slope of the given line can be used to find the slope of a line perpendicular to it.$$\begin{aligned} y = mx+b\end{aligned}$$$$\begin{aligned} y &= \frac{3 x}{2} + 14 \\ \text{Slope} &= m = \frac{3}{2} \end{aligned}$$The negative reciprocal of $\frac{3}{2}$ is $- \frac{2}{3}$, therefore $m$ of a perpendicular line will be $- \frac{2}{3}$.$$\begin{aligned}y &= - \frac{2}{3}x+b \\ 20 &= - \frac{2}{3}(-5)+b \\ b &= \frac{50}{3} \end{aligned}$$The slope-intercept form of the line is: $y = - \frac{2}{3}x +\frac{50}{3}$.