Find the value of "a" for which the graph of the first equation is perpendicular to the graph of the second equation. 1) equation : y=a/3x-6
2) equation 4x+2y=6
their is the same problems but different equation.1) 3y+ax=8
2)3/4x+2

Perpendicular lines will have negative reciprocal slopes. Example : if u r given a slope of 1/3, to find the negative reciprocal, flip the slope and change the sign. So the negative reciprocal of 1/3 is -3/1 or just -3. So slope 1/3 and slope -3 are perpendicular.

equation 2 : 
4x + 2y = 6
2y = -4x + 6
y = -2x + 3...the slope here is -2a perpendicular line will have a slope of 1/2.

y = (a/3)x - 6so we have to make a/3 = 1/21.5/3 = 1/2
so a = 1.5

eq. 2 : 3/4x + 2
not sure what the entire equation isbut it looks to have a slope of 3/4. That means a perpendicular line will have a slope of -4/3.

3y + ax = 9
3y = -ax + 9
y = (-a/3)x + 3therefore, a has to be 4

20y5 times 2 is 1010 times -2 is -20, plus y