MathematicsMedium128×since 2002Q5591The equations of two sides AB\mathrm{AB}AB and AC\mathrm{AC}AC of a triangle ABC\mathrm{ABC}ABC are 4x+y=144 x+y=144x+y=14 and 3x−2y=53 x-2 y=53x−2y=5, respectively. The point (2,−43)\left(2,-\frac{4}{3}\right)(2,−34) divides the third side BC\mathrm{BC}BC internally in the ratio 2:12: 12:1, the equation of the side BC\mathrm{BC}BC isAx+6y+6=0x+6 y+6=0x+6y+6=0Bx−3y−6=0x-3 y-6=0x−3y−6=0Cx+3y+2=0x+3 y+2=0x+3y+2=0Dx−6y−10=0x-6 y-10=0x−6y−10=0Check answerSkip