Find the angle between the straight line 3x-2y=4 and the line joining the points (-2,-1) and (4,1)

Please explain thanks!

slope of 3x - 2y = 4 is 3/2

for the points, change in y is 2, change in x is 6, so slope is 2/3.

slope of a line is tangent of the angle the line makes with the x axis.
3/2 = tan A
2/3 = tan B, so
tan (A - B) = ( 3/2 - 2/3 ) / ( 1 + 3/2•2/3) = (5/6) / 2 = 5/12
so the angle is inverse tangent of 5/12, which is 22.62°

There are a bunch of ways to do this... I'll take a vector approach.

so first we need to points on the line to make a vector with. I'm just going to arbitrarily pick (0, -2) and (1, -1/2)

so v1 is <1 - 0, -1/2 + 2> = <1, 3/2>
v2 = <4 + 2, 1 + 1> = <6, 2>

v1 • v2 = |v1| |v2| cos(theta)

6 + 3 = √(1 + 9/4) √(36 + 4) cos(theta)
9 = √(1 + 9/4) √(40) cos(theta)
9 = √(130)cos(theta)


theta = arccos(9/(√(130)) which is about 0.66104316885 radians

(4,1) (-2,-1)
m=(-1-1)/(-2-4)=-2/-6=1/3
y-1=(1/3)(x-4)

3x-2y=4
-2y=-3x+4
y=(3x/2)-2
m=3/2

arctan((3/2-1/3)/(1+(3/2)(1/3))=0.6610… rad