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 $$=$$

Question

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 $$=$$

Infinity Learn · Accepted Answer

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 $$k=125Therefore$$, $$k=125$$

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

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