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

User Lagoon
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2 Answers

3 votes
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 <==

User Ivan Ivanic
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0 votes
may be
ax+2y+5 = 5x+3y+1
ax-5x = y-4
x(a-5) = y-4
a = (y-4)/x + 5
User Pcasa
by
7.9k points

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