How do you find the slope that is perpendicular to the line #2x +3y = 5#?

Question

Julia Adams · Accepted Answer

Take the negative reciprocal of the given line's slope. The new slope will be #3/2#.

Lines that are perpendicular will have negative reciprocal slopes. Meaning, if one line's slope is #m#, then a perpendicular line will have a slope of #-1/m#.

Why? A line's slope is equal to its rise over its run—also written as #m=[Delta y]/[Delta x]#. If we rotate that line 90° counterclockwise (making it perpendicular to its old self), the run (to the right) becomes a rise (up), and the rise (up) becomes a backwards run (to the left):

thus

#m_"new"=(Delta y_"new")/(Delta x_"new")=(Delta x)/(-Delta y)=-(Delta x)/(Delta y)=-1/m#

(Note: if we rotate this new line another 90° (180° total from the beginning), this 3rd line will have a slope of #(-1)/(-1/m)#, which simplifies to #m#—the same slope of the first line, which is what we would expect.)

Okay, great—so what's the slope of #2x+3y=5#? If we rearrange this into slope-intercept form, we get

#y=-2/3 x+5/3#,
meaning that for every step of "2 down", we have a step of "3 right".

The negative reciprocal of the slope #m=-2/3# is

#m_"new"=-1/m=(-1)/(- 2/3)=3/2#,
meaning that, for a perpendicular line, a step of "3 up" comes with a step of "2 right".

Eva Adamson · Answer

undefined To find the slope that is perpendicular to the line $2x + 3y = 5$, we first need to rearrange the equation into slope-intercept form $y = mx + b$, where $m$ represents the slope.

The slope-intercept form is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

To find the slope of the given line, we can rewrite the equation as $y = \frac{-2}{3}x + \frac{5}{3}$.

The slope of the given line is $\frac{-2}{3}$.

The slope of a line perpendicular to the given line is the negative reciprocal of the slope of the given line.

Therefore, the slope of the line perpendicular to $2x + 3y = 5$ is the negative reciprocal of $\frac{-2}{3}$, which is $\frac{3}{2}$.