Question 217087: find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 3y=6 and passing through (-3,5).
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! base eqn,,,x+3y =6,,,,,change to slope intercept form,,,,3y=-x+6,,,y=(-1/3)x +2,,,,,with m1 = (-1/3) and y intercept = +2
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We know a perpendicular has negative reciprocal slope,,,(m2=-1/m1),,,,therefore the slope of the perpendicular to the base eqn is,,,,,,,,m2= +3
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For the perpendicular line eqn,,,,start with y=mx+b,,,,and y=3x +b
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To find b,,,,,subst ( -3,5) into eqn
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(5)= (+3)(-3) +b
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5=-9=b
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14=b,,,,,,,,subst back into eqn
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y=3x + 14,,,,,,answer to perpendicular line in slope int form
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-14 = 3x-y,,,,,or 3x-y = -14,,,,in standard form
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check, m2= -1/m1,,,,,,,3=-1/(-1/3)= +3,,,,ok
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(-3,5),,,,check fit,,,,,(5) = 3(-3) +14,,,,,,5=-9+14,,,,,,ok
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