Consider the lines x(3 lambda+1)+y(7 lambda+2 | vedclass

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(A) The given equation of the family of lines is $x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5$. Rearranging the terms,we get $(x+2y-5) + \lambda(3x+7y-

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$20$

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(A) The given equation of the family of lines is $x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5$. Rearranging the terms,we get $(x+2y-5) + \lambda(3x+7y-17) = 0$. This family of lines passes through the intersection of $x+2y-5=0$ and $3x+7y-17=0$. Solving these equations: $x = 5-2y$,substituting into the second: $3(5-2y)+7y-17=0 \implies 15-6y+7y-17=0 \implies y=2$. Thus,$x = 5-2(2) = 1$. The fixed point is $P(1, 2)$. The line $L$ that is farthest from the origin $(0, 0)$ is the line perpendicular to the vector $\vec{OP}$,where $O$ is the origin. The slope of $OP$ is $m_{OP} = \frac{2-0}{1-0} = 2$. The slope of line $L$ is $m_L = -\frac{1}{m_{OP}} = -\frac{1}{2}$. The equation of line $L$ passing through $P(1, 2)$ is $y-2 = -\frac{1}{2}(x-1) \implies 2y-4 = -x+1 \implies x+2y-5=0$. We need to find the distance $d$ of line $L$ from the point $Q(3, 6)$. $d = \frac{|1(3) + 2(6) - 5|}{\sqrt{1^2 + 2^2}} = \frac{|3+12-5|}{\sqrt{5}} = \frac{10}{\sqrt{5}} = 2\sqrt{5}$. Therefore,$d^2 = (2\sqrt{5})^2 = 4 	imes 5 = 20$.

Consider the lines $x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5$,where $\lambda$ is a parameter. All these lines pass through a fixed point $P$. One of these lines (say $L$) is farthest from the origin. If the distance of $L$ from the point $(3,6)$ is $d$,then the value of $d^2$ is

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