Find the equation of the line that is perpendicular to the line -5x+2y=1 and passes through the points (-7,4)

asked by guest
on Nov 20, 2024 at 11:58 am



You asked:

Find an equation of the line that passes through the point \(\left(-7, 4\right)\), and is perpendicular to the line with equation \(-\left( 5 x \right) + 2 y = 1\).

MathBot Answer:

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.$$\begin{aligned}Ax + By + C &= 0 \\ -5x +2y -1 &= 0 \end{aligned}$$$$\begin{aligned} \text{Slope} &= -\frac{A}{B} \\ \text{Slope} &= \frac{5}{2} \end{aligned}$$The negative reciprocal of $\frac{5}{2}$ is $- \frac{2}{5}$, therefore $A$ and $B$ of a perpendicular line are $2$ and $5$.$$\begin{aligned} 2x +5y + C &= 0 \\ 2(-7) +5(4)+C &= 0 \\ 20 + C &= 14 \\ C &= -6 \end{aligned}$$An equation of the line in standard form is: $2x +5y -6 = 0$.


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.$$\begin{aligned} y = mx+b\end{aligned}$$$$\begin{aligned} y &= \frac{5 x}{2} + \frac{1}{2} \\ \text{Slope} &= m = \frac{5}{2} \end{aligned}$$The negative reciprocal of $\frac{5}{2}$ is $- \frac{2}{5}$, therefore $m$ of a perpendicular line will be $- \frac{2}{5}$.$$\begin{aligned}y &= - \frac{2}{5}x+b \\ 4 &= - \frac{2}{5}(-7)+b \\ b &= \frac{6}{5} \end{aligned}$$The slope-intercept form of the line is: $y = - \frac{2}{5}x +\frac{6}{5}$.