The graphs of $2y + x + 3 = 0$ and $3y + ax + 2 = 0$ are perpendicular. Solve for $a.$

VoidAsks Apr 17, 2021

#2**+3 **

If they are perpindicular , then the slope, m of line 1 will be - 1/m of the second line...

re-arrange the equations into y = mx + b form to easily see the slopes

2y + x + 3 = 0 becomes

y = - 1/2 x - 3/2 slope , m = - 1/2 does the slope of the other line = - 1/ (-1/2) = 2 ???

second line

3y + ax + 2 = 0

3y= - ax - 2

y = - a/3 x - 2/3 to be perpindicular - a/3 must equal 2

- a/3 = 2

** a = -6 **

ElectricPavlov Apr 17, 2021

#2**+3 **

Best Answer

If they are perpindicular , then the slope, m of line 1 will be - 1/m of the second line...

re-arrange the equations into y = mx + b form to easily see the slopes

2y + x + 3 = 0 becomes

y = - 1/2 x - 3/2 slope , m = - 1/2 does the slope of the other line = - 1/ (-1/2) = 2 ???

second line

3y + ax + 2 = 0

3y= - ax - 2

y = - a/3 x - 2/3 to be perpindicular - a/3 must equal 2

- a/3 = 2

** a = -6 **

ElectricPavlov Apr 17, 2021