How you find the value of m so that the lines with equations #-3y+2x=4# and #mx + 2y = 3# are perpendicular?

Question

Julia Adams · Accepted Answer

#m=3#

Put both equations into the form #y=mx+c#

#-3y+2x=4#

#3y=2x-4#

#y=2/3x-4/3#

Regarding the second equation:

#mx+2y=3#

#2y=3-mx#

#y=-m/2x-3/2#

The product of the gradients of the lines is -1 if they are perpendicular.

So #-m/2xx2/3=-1#

#m=3#

Savannah Anderson · Answer

undefined To find the value of $m$ so that the lines with equations $-3y + 2x = 4$ and $mx + 2y = 3$ are perpendicular, you need to ensure that the slopes of the two lines are negative reciprocals of each other.

The first equation is in the form $Ax + By = C$, where $A = 2$, $B = -3$, and $C = 4$. The slope of this line can be found by rearranging the equation to solve for $y$, giving $y = \frac{2}{3}x - \frac{4}{3}$. So, the slope of the first line is $\frac{2}{3}$.

For the second equation $mx + 2y = 3$, rearrange it to solve for $y$, which becomes $y = -\frac{m}{2}x + \frac{3}{2}$. Therefore, the slope of the second line is $-\frac{m}{2}$.

To make the lines perpendicular, the product of their slopes must be -1:

$$\left(\frac{2}{3}\right) \times \left(-\frac{m}{2}\right) = -1$$

Solving this equation for $m$ will give the required value:

$$\frac{2}{3} \times \left(-\frac{m}{2}\right) = -1$$

$$ \frac{-m}{3} = -1$$

$$ -m = -3$$

$$ m = 3$$

So, for the lines to be perpendicular, $m$ must equal $3$.