The line \(x+y=1\) meets the lines represented by the equation \(y^3-6 x y^2+11 x^2 y-6 x^3=0\) at the…
The line \(x+y=1\) meets the lines represented by the equation \(y^3-6 x y^2+11 x^2 y-6 x^3=0\) at the points \(P, Q, R\). If \(O\) is the origin, then \((O P)^2+(O Q)^2+(O R)^2=\)
\(\frac{85}{72}\)
\(\frac{121}{72}\)
\(\frac{212}{72}\)
\(\frac{217}{72}\)
Solution
We are given with a cubic equation which represents 3-lines.
Let lines have,
\(y-m_1 x=0 ; y-m_2 x=0\)
and \(y-m_3 x=0\)
Then, \(\left(y-m_1 x\right)\left(y-m_2 x\right)\left(y-m_3 x\right)=0\)
is the combined equations
Which in this case is,
\(y^3-6 x y^2+11 x^2 y-6 x^3=0\)
By comparison we are able to put above equation are,
\((y-x)(y-2 x)(y-3 x)=0\)
So, lines are,
$\begin{aligned}
L_1 & \Rightarrow y-x=0 \\
L_2 & \Rightarrow y-2x=0 \\
\text{and } L_3 & \Rightarrow y-3x=0
\end{aligned}$
Their intersection points with
$\begin{aligned}
x+y=1 & \text{ are, } \\
P & \equiv x+y=1 \text{ and } y-x=0 \\
\Rightarrow \quad P & \equiv\left(\frac{1}{2}, \frac{1}{2}\right) \\
Q & \equiv x+y=1 \text{ and } y-2x=0
\end{aligned}$
$\begin{aligned}
& \Rightarrow \quad Q \equiv\left(\frac{1}{3}, \frac{2}{3}\right) \\
& R \equiv x+y=1 \text{ and } y-3x=0 \\
& \Rightarrow \quad R \equiv\left(\frac{1}{4}, \frac{3}{4}\right) \\
& \text{So, } Q P^2+O Q^2+O R^2 \\
&=\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{9}+\frac{4}{9}\right)+\left(\frac{1}{16}+\frac{9}{16}\right) \\
&=\frac{2}{4}+\frac{5}{9}+\frac{10}{16}=\frac{36+40+45}{72}=\frac{121}{72}
\end{aligned}$