AQ3.3: Activity Questions 3

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pradeepa...@gmail.com

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Nov 4, 2020, 10:10:41 AM11/4/20
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The equation of a line passing through the point (3,4) and perpendicular to the line 3x+4y8=0 is
How come to this answer 
8x6y=0

Anand Iyer

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Nov 4, 2020, 10:11:06 PM11/4/20
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the slope of the given line 3x+4y−8=0  is -3/4

Hence, the slope of the perpendicular is 4/3.

y = (4/3)x + c is the equation of line we need.

In order to find c, substitute the point (3, 4) into the above equation.

4  = (4/3) * 3 + c ==> c = 0

Thus, equation of the line we require is y = (4/3) * x ==> 4x - 3y = 0.  Multiplying 2 on both sides of the equation, we get 8x - 6y = 0

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