Question 322079: Find an equation of the line containing the point (1,-2) and parallel to the line 9x-3y=5. Show work and explain each step. Please help!!!!!!!Don't not understand...Thanks
Found 2 solutions by stanbon, Jk22:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Find an equation of the line containing the point (1,-2) and parallel to the line 9x-3y=5. Show work and explain each step.
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Given line: 9x-3y = 5
Slope-int form: y = 3x - (5/3)
Note: slope = 3
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Find Equation of line with slope = 3, passing thru (1,-2)
Form: y = mx + b
Substitute:
-2 = 3(1) = b
b = -5
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Equation:
y = 3x - 5
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Cheers,
Stan H.
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Answer by Jk22(389)  (Show Source):
You can put this solution on YOUR website! A line is given by : set of (x,y)=OP+a*(dx,dy)
OP is a point of the line, (dx,dy) is the vector giving the direction of the line.
we need 2 points of the line : e.g. OP=(0,-5/3) and OP1=(1,4/3)
hence : (dx,dy)=(1,3)
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For the seeked line we put OP2=(1,-2) and the same vector of direction, to be parallel :
line 2 is the set of :(x2,y2)=(1,-2)+a(1,3)
x2=1+a
y2=-2+3a
eliminate a : 3x2-y2=5
hence : equation is : 9x-3y=15 (which is just first equation, with the RHS
obtained by putting (x,y)=(1,-2) : 9-3*(-2)=15)
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