Math Problem Solver

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.$$\begin{aligned} y = mx+b\end{aligned}$$$$\begin{aligned} y &= - \frac{2 x}{3} - \frac{1}{3} \\ \text{Slope} &= m = - \frac{2}{3} \end{aligned}$$The negative reciprocal of $- \frac{2}{3}$ is $\frac{3}{2}$, therefore $m$ of a perpendicular line will be $\frac{3}{2}$.$$\begin{aligned}y &= \frac{3}{2}x+b \\ 2 &= \frac{3}{2}(1)+b \\ b &= \frac{1}{2} \end{aligned}$$An equation of the line in slope-intercept form is: $y = \frac{3}{2}x +\frac{1}{2}$.

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Point-Slope 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.$$\begin{aligned} y &= - \frac{2 x}{3} - \frac{1}{3} \\ \text{Slope} &= m = - \frac{2}{3} \end{aligned}$$The negative reciprocal of $- \frac{2}{3}$ is $\frac{3}{2}$, therefore $m$ of a perpendicular line will be $\frac{3}{2}$.$$\begin{aligned} y - y_{1} = m(x - x_{1}) \end{aligned}$$$$\text{where m } = \frac{3}{2}, \text{ } x_{1} = 1, \text{ and } y_{1} = 2$$An equation of the line in point-slope form is: $y - 2 = \frac{3 \left(x - 1\right)}{2}$

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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.$$\begin{aligned}Ax + By + C &= 0 \\ 2x +3y +1 &= 0 \end{aligned}$$$$\begin{aligned} \text{Slope} &= -\frac{A}{B} \\ \text{Slope} &= - \frac{2}{3} \end{aligned}$$The negative reciprocal of $- \frac{2}{3}$ is $\frac{3}{2}$, therefore $A$ and $B$ of a perpendicular line are $3$ and $-2$.$$\begin{aligned} 3x -2y + C &= 0 \\ 3(1) -2(2)+C &= 0 \\ 3 + C &= 4 \\ C &= 1 \end{aligned}$$An equation of the line in standard form is: $3x -2y +1 = 0$