Question 464989: ok here's the question "find the equation in standard form and in slope intercept form and y-intercept form' and graph of a line
a-have an x-intercept (3,0) and a slope of -2/3
b-containing the points (-8,-1) (-5,9)
C- passes through the point (0,-3) and is parrallel to 4x+3y=-1
d- passes through the point (6,-2) and is perpindecular to y=-1/3x+2
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! "find the equation in standard form and in slope intercept form and y-intercept form' and graph of a line
a-have an x-intercept (3,0) and a slope of -2/3
Form: y = mx+b
Substitute for x,y,m and solve for "b":
0 = (-2/3)*3 + b
b = -2
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Equation: y = (-2/3)x-2
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b-containing the points (-8,-1) (-5,9)
m = (9--1)/(-5--8) = 10/3
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Using (-5,9) and m = 10/3, solve for "b":
9 = (10/3)(-5) + b
(27/3) = (-50/3) + b
b = (70/3)
Equation:
y = (10/3)x+(70/3)
=========================================
C- passes through the point (0,-3) and is parallel to 4x+3y=-1
The given line has slope -4/3
The y-intercept = -3
Eqaution:
y = (-4/3)x-3
=========================================
d- passes through the point (6,-2) and is perpindecular to y=-1/3x+2
The given line has slope = (-1/3)
The desired slope is m = 3
---
Using x,y,m solve for "b":
-2 = 3*6 + b
b = -20
----
Equation:
y = 3x-20
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Cheers,
Stan H.
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