Question

x-3y+6=0, 2x+3y+10=0 find the angle between them​

Answer 1

Answer:

52 degree.

Step-by-step explanation:

If the two lines are a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, then the formula to find the angle between them is

tan θ = |(a1b2 - b1a2)/(a1a2 + b1b2)|

x - 3y +6 = 0 ..........(1)

Then a1 = 1, b1 = -3, c1 = 6

2x + 3y +10 = +0 ..............(2)

Then a2 = 2, b2 = 3, c2 = 10

tan θ = |(1*(3) -(-3)*2)/(1*2 + 3*(-3))|

tan θ = |3+6/2-9| =9/7

taking the inverse of tangent, we get

So, angle θ = 52.125 degree.

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