Question 201322: Find an equation of the line satisfying the given conditions.
Through (-4,3); perpendicular to -8x+5y=77
Found 2 solutions by Earlsdon, RAY100: Answer by Earlsdon(6294) (Show Source): Answer by RAY100(1637) (Show Source):
You can put this solution on YOUR website! For this problem, lets first find the slope of the original equation, using y=mx +b form, and then define the slope of the second eqn as, m2=-1/m1,perpendicular. Finally, we can define the eqn of the second line from the slope and the given point.
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-8x +5y =77
5y = 8x +77
y = (8/5)x + 77/5
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or from form ,,,,y= m x +b,,,m = (8/5)
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m2= - 1/m1 = (-5/8)
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Second eqn is , y= mx +b,,,,y = (-5/8) x +b
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As it contains the point ( -4,3),,,,substitute
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(3) = (-5/8) (-4) + b
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3 = 20/8 +b
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24/8 - 20/8 = 4/8 =1/2 = b
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subst m,,and b,,in form eqn,, y = (-5/8)x + (1/2)
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mult thru by 8
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8y = -5x + 4
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checking, subst ( -4,3),,,,8*3=(-5)(-4) +4 = 24 ,,,,ok
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