Question 1129600
Write the equation of the line that satisfies the given conditions. Express the final equation in standard form. 
Contains the point (-3, 5) and is PARALLEL to the line x − 5y = 6
Solve the equation for "y"::
5y = x-6
y = (1/5)x-(6/5)
The slope is 1/5
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Equation of line with slope = 1/5 passing thru point (-3,5)
(y-5)/(x+3) = 1/5
x+3 = 5y-25
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Standard Form:: x-5y = -28
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Contains the point (-4, 7) and is PERPENDICULAR to the line 3x − y = 4 
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Find the slope of the given equation::
y = 3x-4
slope = 3
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Equation we want must have slope = -1/3 and must pass thru (-4,7)
y-7 = (-1/3)(x+4)
Find the standard form::
x+4 = -3(y-7)
x+4 = -3y+21
Ans:: x + 3y = 17
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Cheers,
Stan H.
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