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

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

Apr 17, 2021

#1
0

A is... Never mind you got someone else that is about to answer :/

Bye~Hannah

Apr 17, 2021
#3
+1

WillBillDillPickle  Apr 17, 2021
#4
0

Cant I've tried it wont send me an email

Guest 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