The product of perpendicular from (1,1) to the pair of lines 2x2+3xy+y2=0is equal to the product of perpendiculars from (0,0) to the pair of lines x2+2xy+3y2+x+2y+k=0 then k =

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512

−125

125

−512

answer is C.

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Detailed Solution

The product of the perpendiculars drawn from the point Px1,y1to the pair of lines represented by the equation ax2+2hxy+by2=0 is ax12+2hx1y1+by12a−b2+4h2The product of the perpendiculars drawn from the point P0,0to the pair of lines represented by the equation ax2+2hxy+by2+2gx+2fy+c=0 is ca−b2+4h2Given that product of perpendicular from (1,1) to the pair of lines 2x2+3xy+y2=0is equal to the product of perpendiculars from (0,0) to the pair of lines x2+2xy+3y2+x+2y+k=0Hence, 212+311+122−12+32=k1−32+2h2610=k8Cross multiply and then simplify k=125Therefore, k=125

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