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