The line $L_{1}$ : 4x + 3y – 12 = 0 intersect the x – axis and y – axis at A and B respectively. A variable line perpendicular to $L_{1}$ intersects the x & y – axis at ‘P’ & ‘Q’ respectively then the locus of the circumcentre of triangle ABQ is

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6x – 8y + 7 = 0

6x + 8y – 25 = 0

8x – 6y + 7 = 0

14x – 6y + 7 = 0

answer is A.

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Detailed Solution

Circumcentre of a triangle ABQ will lie on the perpendicular bisector of line AB. Now equation of perpendicular bisector of line AB is $3x- 4y + \frac{7}{2} = 0$ . Hence locus of circumcentre is 6x – 8y + 7 = 0.

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