If the line ax+2y+5=0 and 5x+3y+1=0, find the value of a

5x + 3y + 1 = 0

3y = - 5x - 1

y = - 5/3x - 1/3 ... slope here is - 5/3. perpendicular lines will have negative reciprocal slopes. So the other line will have to have a slope of 3/5.

ax + 2y + 5 = 0

2y = - ax - 5

y = - 1/2ax - 5/2

-1/2a = 3/5

-a = 3/5 * 2

-a = 6/5

a = - 6/5

lets check this ... it looks funny

-6/5x + 2y + 5 = 0

2y = 6/5x - 5

y = (6/5x * 1/2) - 5/2

y = 3/5x - 5/2 ... and the slope is 3/5

ok, then ur answer is : a = - 6/5 <==

May be

ax+2y+5 = 5x+3y+1

ax-5x = y-4

x (a-5) = y-4

a = (y-4) / x + 5