2x-3y-6=0 has slope 2/3
x+5y-20=0 has slope -1/5
so, you want
arctan(-1/5) = -11.3°
arctan(2/3) = 33.7°
so, the acute angle between the lines is 45°
Find the acute angle between the lines 2x-3y-6=0 and x+5y-20=0
3 answers
You should know that the slope of a line is the same as the tangent of the angle that the line makes with the x-axis
2x-3y-6=0 , slope = 2/3,
tanØ = 2/3,
Ø = appr 33.69° with the x-axis
same with x+5y-20=0
slope = -1/5 , line leans to the left.
angle with x-axis is appr 168.69°
angle between the lines = 168.69 - 33.69 = 135°
or, if Ø is the angle between them
tanØ = (m2 - m1) /( 1 + m2m1)
= (-1/5 - 2/3) / (1 + (-1/5)(2/3)
= (-13/15) / (13/15)
= -1
then Ø = arctan(-1) = 135°
2x-3y-6=0 , slope = 2/3,
tanØ = 2/3,
Ø = appr 33.69° with the x-axis
same with x+5y-20=0
slope = -1/5 , line leans to the left.
angle with x-axis is appr 168.69°
angle between the lines = 168.69 - 33.69 = 135°
or, if Ø is the angle between them
tanØ = (m2 - m1) /( 1 + m2m1)
= (-1/5 - 2/3) / (1 + (-1/5)(2/3)
= (-13/15) / (13/15)
= -1
then Ø = arctan(-1) = 135°
Steve's answer of 45° is correct, I didn't see the "acute" angle part, and 180-135 = 45°