find angle between two lines

Question

Find the angle between two lines

3x-5y=3

3x+5y=12

Answer 1

The line equations are 3x -5y = 3 and 3x +5y = 12.

3x-5y -3x = 3-3x                                  (Subtract 3x from each side )

-5y = -3x + 3

y = -(3/-5)x - 3/-5                                  (Divide each side by -5)

y = (3/5)x +3/5                                 

Compared to the slope intercept equation y = mx +b

slope(m₁) = 3/5, and intercept(b₁) = 3/5

And again the next equation Write the line equation in slope-intercept form y = mx + b, m is slope and b is y-intercept.

3x +5y = 12

3x +5y -3x = 12 -3x                            (Add -3x to each side )

5y = 12 - 3x    

5y/5 = (-3x+12)/5                               (Divide  each side by 2)

y = -3x/5 +12/5                                                   

Compared to the slope intercept equation y = mx +b

slope(m₂) = 3/5, and intercept(b₂) = 12/5

The Angle between two lines 

tan(Θ) = |(m₁ -m₂)/(1+m₁m₂)|

Substitute the value of m₁ & m₂ above equation

tan(Θ) = ((3/5) - (-3/5))/(1+(3/5) (-3/5))   (LCM of 3/5 +3/5= 6/5)    

tan(Θ) = 6/5/(1-9/25)                        (LCM of 25,1= 25)                    

tan(Θ) = 6/5/((25-9)/25)                                   

tan(Θ) =  6/5/((16)/25)                      (Simplify)             

tan(Θ) = 15/8                                           

The value of (Θ) = tan^-1 (15/8).

Comments

tan(θ) = | (m₁ -m₂)/(1+m₁m₂) |.

θ = tan^-1 (15/8) = 61.93^O.

Therefore, the angle between lines (θ) is 61.93^O .