JEE Main 202313 Apr, Evening shiftMathematicsHardCoordinate Geometry478× since 2002Let (α,β)(\alpha, \beta)(α,β) be the centroid of the triangle formed by the lines 15x−y=82,6x−5y=−415 x-y=82,6 x-5 y=-415x−y=82,6x−5y=−4 and 9x+4y=179 x+4 y=179x+4y=17. Then α+2β\alpha+2 \betaα+2β and 2α−β2 \alpha-\beta2α−β are the roots of the equation :AOption A. x2−7x+12=0x^{2}-7 x+12=0x2−7x+12=0BOption B. x2−13x+42=0x^{2}-13 x+42=0x2−13x+42=0COption C. x2−14x+48=0x^{2}-14 x+48=0x2−14x+48=0DOption D. x2−10x+25=0x^{2}-10 x+25=0x2−10x+25=0CheckSkip · show solution
