The line x+y=1 meets the lines represented by | vedclass

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(B) The given equation is $y^3-6xy^2+11x^2y-6x^3=0$. Factoring the cubic expression,we get $(y-x)(y-2x)(y-3x)=0$. Thus,the three lines are $L_1: y=x$,

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$\frac{121}{72}$

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(B) The given equation is $y^3-6xy^2+11x^2y-6x^3=0$. Factoring the cubic expression,we get $(y-x)(y-2x)(y-3x)=0$. Thus,the three lines are $L_1: y=x$,$L_2: y=2x$,and $L_3: y=3x$. We find the intersection points of these lines with $x+y=1$: For $P$ $(y=x)$: $x+x=1 \Rightarrow x=1/2, y=1/2$. So $P = (1/2, 1/2)$. For $Q$ $(y=2x)$: $x+2x=1 \Rightarrow x=1/3, y=2/3$. So $Q = (1/3, 2/3)$. For $R$ $(y=3x)$: $x+3x=1 \Rightarrow x=1/4, y=3/4$. So $R = (1/4, 3/4)$. Now,calculate the squared distances from the origin $O(0,0)$: $(OP)^2 = (1/2)^2 + (1/2)^2 = 1/4 + 1/4 = 1/2 = 36/72$. $(OQ)^2 = (1/3)^2 + (2/3)^2 = 1/9 + 4/9 = 5/9 = 40/72$. $(OR)^2 = (1/4)^2 + (3/4)^2 = 1/16 + 9/16 = 10/16 = 5/8 = 45/72$. Summing these: $(OP)^2+(OQ)^2+(OR)^2 = 36/72 + 40/72 + 45/72 = 121/72$.

The line $x+y=1$ meets the lines represented by the equation $y^3-6xy^2+11x^2y-6x^3=0$ at the points $P, Q, R$. If $O$ is the origin,then $(OP)^2+(OQ)^2+(OR)^2=$

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