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Question

The equation of a straight line, perpendicular to 3x−4y=6 and forming a triangle of area 6 square units with coordinate axes, is

A
x2y=6
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B
4x+3y=12
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C
4x+3y+24=0
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D
3x+4y=12
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Solution

The correct option is D 4x+3y=12
The line which is perpendicular to the given line will have a slope of 43

The equation of the line will look like 3y+4x=k

This line intersects the x-axis at (k4,0) an y-axis at (0,k3).

The area of the triangle formed is given by 12×k3×k4=6

k2=144

k=±12

Thus, the line equation becomes 4x+3y=±12

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