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Offline Centres

Q.

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

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a

3x−4y+2=0

b

4x+3y+7=0

c

6x−8y+7=0

d

4x−3y+7=0

answer is C.

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Detailed Solution

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

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