Question

The line $x+y=1$ meets the lines represented by the equation $y^3-x{y}^2-14x^{2}y+24x^3=0$ at the points $A,B,C$. If $O$ is the origin, then $O{A}^2+O{B}^2+O{C}^2$ is equal to

• A. $\frac{22}{9}$
• B. $\frac{85}{72}$
• C. $\frac{181}{72}$
• D. $\frac{221}{72}$

Answer 1

The correct option is D $\frac{221}{72}$

X-coordinate of the points are given by the roots of the equation

$24x^3+14x^2(x-1)-x{(x-1)}^2-{(x-1)}^3=0\Rightarrow 36x^3-9x^2-4x+1=0\Rightarrow (3x-1)(3x+1)(4x-1)=0\Rightarrow x=\frac{1}{3},-\frac{1}{3},\frac{1}{4}$

$\Rightarrow A(\frac{1}{3},\frac{2}{3}),B(-\frac{1}{3},\frac{4}{3})$ and $C(\frac{1}{4},\frac{3}{4})$

Hence,

${OA}^2+{OB}^2+{OC}^2=\frac{1}{9}+\frac{4}{9}+\frac{1}{9}+\frac{16}{9}+\frac{1}{16}+\frac{9}{16}=\frac{221}{72}$