SOLUTION: Find the point of intersection and the angle between 2x-y+3=0 and 3x+y-1=0. Use decimal values for fractional answers.

Question 1106127: Find the point of intersection and the angle between 2x-y+3=0 and 3x+y-1=0. Use decimal values for fractional answers.
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
1) 2x-y+3=0
:
2) 3x+y-1=0
:
solve equation 1 for y
:
3) y = 2x+3
:
solve equation 2 for y
:
4) y = -3x+1
:
set equations 3 and 4 equalt to each other and solve for x
:
2x+3 = -3x+1
:
5x = -2
:
x = -2/5 = -0.4
:
use equation 3, substitute for x
:
y = 2(-2/5) +3 = 2.2
:
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point of intersection is (-0.4,2.2)
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we use the slopes of the lines to find the angle between them
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slope of equation 1 is 2 and the slope of equation 2 is -3
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the angle between the two lines is the inverse tangent (m1 - m2)/(1 + m1*m2)
where m1 is the slope of line 1 and m2 is the slope of line 2
:
tan^(-1) = (2 - (-3)) / (1 + 2*(-3)) = tan^(-1) (-5/5) = -45 degrees
:
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the angle between the two lines is measured counter clockwise so the angle between the two lines is 45 degrees
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