# What is the angle between the lines x + y = 3 and x + 3y = 5?

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We need to find the angle between the lines x + y = 3 and x + 3y = 5.

Now we first write the two lines in the form y = mx +c where m is the slope of the line.

x + y = 3

=> y = -x + 3

The slope is -1.

x + 3y = 5

=> 3y = -x + 5

=> y = -x/3 + 5/3

The slope is -1/3

If the angle between two lines given the slopes m1 and m2 is x, we have tan x = (m2 – m1)/( 1+ m1*m2)

So tan x = (-1 + 1/3)/( 1+ 1/3)

=> tan x = (-2/3)/(4/3)

=> tan x = (-2/4)

=> tan x = -1/2

x = arc tan (-1/2)

x = 26.56 degrees

**Therefore the angle between the lines is 26.56 degrees.**

x+y = 3. and x+3y = 5.

We write both lines in the slope intercept form like y = mx+c, where m is the slope (tangent of the angle with x axis) of the line and c is intercept of the line on y axis.

x+y = 3 is written like: y = -x+3.....(1)

x+3y = 5 is written in the form y = -(1/3)x+5/3....(2)

So the slope of the first line m1 = -1. and the 2nd line = m2 -1/3.

Therefore the angle between the lines is given by:

tanA = (m1-m2)/(1+m1m2) = (-1+1/3)/{1+(-1)(-1/3)}

tan A = {(-2/3)/((4/3)} = -2/4 = -1/2.

Therefore A = arc tan (-1/2) = -26.56 degre from line (1) to line (2).

Or 26. 56 degree from line (2) to line (1).