The line L1=4x+3y−12=0 intersects the x-and y-axis A and B, respectively. A variable line perpendicular to L1 intersects the x-and the y-axis at P and Q, respectively. Then the locus of the circumcentre of triangle ABQ is

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3x−4y+2=0

4x+3y+7=0

6x−8y+7=0

none of these

answer is C.

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Detailed Solution

Clearly, the circumcenter of triangle ABQ will lie on the perpendicular bisector of line AB. Now, the equations of perpendicular bisector of line AB is 3x−4y+7/2=0. Hence, the locus of circumcentre is 6x−8y+7=0

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