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Question 1198524: Write an equation (a) in slope-intercept form and (b) in standard form for the line passing through (-2,6) and parallel to x+3y=5.
Found 3 solutions by MathLover1, greenestamps, math_tutor2020:
Answer by MathLover1(20850)   (Show Source):
Answer by greenestamps(13214)  (Show Source):
You can put this solution on YOUR website!
The question asks for an answer in both slope-intercept form and standard form, and the given equation is in standard form. So the fastest path to the solution is to find the equation of the parallel line first in standard form.
Given equation: x+3y=5
Equation of parallel line through (-2,6): -2+3(6)=16, so x+3y=16
Convert to slope-intercept form:
x+3y=16
3y=-x+16
y=(-1/3)x+16/3
ANSWERS: y=(-1/3)x+16/3; x+3y=16
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
This might seem strange, but I'll do part (b) first.
Then I'll handle part (a).
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Part (b)
A linear equation is in standard form if it matches with the template Ax+By = C.
Some textbooks use Ax+By+C = 0, which is the same thing more or less, but I'll go with the first template.
The A,B,C values are integers.
Comparing
x+3y = 5
to
Ax+By = C
shows us that
A = 1
B = 3
C = 5
Anything parallel to Ax+By = C is of the form Ax+By = D, where D is also an integer, and it cannot equal C.
Let's use the coordinates of the point (x,y) = (-2,6) to find the value of D.
Ax+By = D
1x+3y = D
x+3y = D
-2+3(6) = D
-2+18 = D
16 = D
D = 16
The standard form equation for the parallel line passing through (-2,6) is x+3y = 16
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Part (a)
Solve the result of part (b) for y.
x+3y = 16
3y = -x+16
y = (-x+16)/3
y = -x/3 + 16/3
y = (-1/3)x + 16/3
Answer: y = (-1/3)x + 16/3
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