Find the acute angle between the lines 2x-3y-6=0 and x+5y-20=0
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°
Steve's answer of 45° is correct, I didn't see the "acute" angle part, and 180-135 = 45°
To find the acute angle between two lines, you need to determine the slopes of the lines and then apply the formula:
tan(acute angle) = |(m1 - m2)/(1 + m1 * m2)|
where m1 and m2 are the slopes of the lines.
First, let's put the given equations in slope-intercept form (y = mx + b):
2x - 3y - 6 = 0 --> 3y = 2x - 6 --> y = (2/3)x - 2
x + 5y - 20 = 0 --> 5y = -x + 20 --> y = (-1/5)x + 4
The slope of the first line (m1) is 2/3, and the slope of the second line (m2) is -1/5.
Now, substitute these values into the formula:
tan(acute angle) = |(m1 - m2)/(1 + m1 * m2)|
= |((2/3) - (-1/5))/(1 + (2/3) * (-1/5))|
= |((2/3) + (1/5))/(1 - (2/15))|
= |[(10 + 3)/(15 - 2)] / (13/15)|
= |(13/15) / (13/15)|
= 1
Since the tangent of an acute angle is always positive, the angle between the lines is 45 degrees or π/4 radians.
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°