Question

find the angle between two straight line y = 3x+8 and 3y-x-9=0​

Answer 1

Answer:

Given equations are,

3x -y + 8 = 0 and -x + 3y - 9 = 0

So, a1 = 3, b1 = -1, a2 = -1, b2 = 3

to find angle we have,

tan θ = |(a1b2 - b1a2)/(a1a2 + b1b2)|

tan θ = | [3(3) - (-1)(-1)] / [3(-1) + (-1)3] |

tan θ = | [8] / [-6] | = 4/3

θ = 53.12