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Question

What is the equation of the line which passes through (4,−5) and is perpendicular to 3x+4y+5=0?

A
4x3y31=0
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B
3x4y41=0
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C
4x+3y1=0
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D
3x+4y+8=0
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Solution

The correct option is C 4x3y31=0
Slope of the given line 3x+4y+5=0 is m1=34
Slope of the line perpendicular to the given line is 1m1=1(34)=43

The equation of the line which is perpendicular to 3x+4y+5=0 and passing through point (4,5) is yy1=m(xx1)
y(5)=43(x4)3y+15=4x164x3y31=0

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