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What is the equation of the line, in general form, that passes through (1, 2) and is parallel to the line whose equation is 3x - 4y + 12 = 0?
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So first let's rewrite the line: 3x-4y+12=0 -4y = -12-3x 4y = 12+3x y = (3/4)x + 3
ok got that
Now if the lines are parallel the slopes must be the same, so the sloep is: M = 3/4
ok is that it?
Not yet, now find the equation of the line you need: y-y1 = m(x-x1) y - 2 = (3/4)(x-1)
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y - 2 = (3x/4) - (3/4) y = (3/4)x - (3/4) + 2 y = (3/4)x - (3/4) + (8/4)
y = (3/4)x + (5/4) or y = (1/4) (3x + 5)
or 4y = 3x + 5, or 4y-3x-5=0.. whichever form you want.. all of the last 4 lines are the same..
thanksss so ,much and tahank you for helpin me!
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