Find the acute angle between the lines y=2x+4 and y=-3x+6

Find the acute angle between the lines y=2x+4 and y=-3x+6

Let the lines' angles of elevation be α and β, respectively. The slope of each line is the tangent of its angle of elevation.

tanα = 2
tanβ = -3

Use the tangent of difference formula to find the difference of the angles.

tan(α - β) = (tanα - tanβ) / [1 + (tanα)(tanβ)]
= [2 - (-3)] / [1 + (2)(-3)]
= -1

tan⁻¹(-1) = -45°

The acute angle of intersection is 45°.

1) The acute angle between two lines is givne by:

θ = tan⁻¹|(m2 - m1)/{1 + (m1)*(m2)|

2) Here, m1, which is the slope of the first line = 2 and m2, which is the slope of the second line = -3

Substituting in the above formula,

θ = tan⁻¹|(-3-2)/(1+6)|

==> θ = tan⁻¹(5/7)

y=2x+4 .........(i)
m1 = 2
and y=-3x+6 ...............(ii)
m2 = 3
using the formula
tan(theta) = (+/-)[(m1 - m2)/(1+m1*m2)
= +/- (2 -3)/(1+2*3)
= (+/-)(-1/7)
tan (theta) = 1/7 take +ve sign for acute angle
theta = tan^-1(1/7) = 8.13010235415575 degrees ..............Ans