Question 824658: Find the equation of the line which passes through the point of intersection of 5x-3y+2=0 and x+3y+1=0 and is parrallel to 2x+3y+3=0. pleasseee answer
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! ---
5x-3y+2=0
x+3y+1=0
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put the system of linear equations into standard form
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5x - 3y = -2
x + 3y = -1
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copy and paste the above standard form linear equations in to this solver:
https://sooeet.com/math/system-of-linear-equations-solver.php
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point of intersection:
x= -0.5 = -1/2
y= -0.1666666 = -1/6
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(given) our new line is parallel to:
2x + 3y + 3 = 0
in slope intercept form:
3y = -2x - 3
y = (-2/3)x - 1
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slope of our new parallel line = -2/3 = -0.666666
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to solve the linear equation, copy and paste this (the point of intersection and the parallel slope):
-0.5, -0.1666666, -0.666666
into the "Point-Slope form: x1 y1 m" input box here: https://sooeet.com/math/linear-equation-solver.php
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x-intercept = (-0.75, 0)
y-intercept = (0, -0.5)
slope = -0.666666
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answer:
the slope-intercept form of our new line:
y = (-0.666666)x - 0.5
y = (-2/3)x - 1/2
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