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

### Related Formula A family of lines passing through the intersection of L_1 = 0$L_1 = 0$ and L_2 = 0$L_2 = 0$ is expressed as: L_1 + lambda L_2 = 0$$L_1 + \lambda L_2 = 0$$ For a point P$P$ through which a family of lines passes, the line in the family that is at the maximum distance from origin O$O$ is the line perpendicular to OP$OP$ passing through P$P$. ### Core Logic Rearranging the equation of the given lines in terms of lambda$\lambda$: (x + 2y - 5) + lambda(3x + 7y - 17) = 0$$(x + 2y - 5) + \lambda(3x + 7y - 17) = 0$$ To find point P$P$, solve the system: 1. x + 2y = 5 implies x = 5 - 2y$x + 2y = 5 \implies x = 5 - 2y$ 2. 3x + 7y = 17$3x + 7y = 17$ ### Step 1: Finding P$P$ and Line L$L$ Substitute x = 5-2y$x = 5-2y$ into the second equation: 3(5 - 2y) + 7y = 17 implies 15 + y = 17 implies y = 2$$3(5 - 2y) + 7y = 17 \implies 15 + y = 17 \implies y = 2$$ x = 5 - 2(2) = 1$$x = 5 - 2(2) = 1$$ Thus, the common intersection point is P(1, 2)$P(1, 2)$. The line farthest from the origin is perpendicular to the segment OP$OP$ joining the origin O(0,0)$O(0,0)$ to P(1,2)$P(1,2)$. - Slope of OP = frac2 - 01 - 0 = 2$OP = \frac{2 - 0}{1 - 0} = 2$ - Slope of L$L$ (m$m$) = -frac12$= -\frac{1}{2}$ Equation of line L$L$ passing through P(1,2)$P(1,2)$: y - 2 = -frac12(x - 1) implies 2y - 4 = -x + 1 implies x + 2y - 5 = 0$$y - 2 = -\frac{1}{2}(x - 1) \implies 2y - 4 = -x + 1 \implies x + 2y - 5 = 0$$ ### Step 2: Distance calculation from (3,6)$(3,6)$ The distance d$d$ of point Q(3,6)$Q(3,6)$ from line x + 2y - 5 = 0$x + 2y - 5 = 0$ is: d = left| frac3 + 2(6) - 5sqrt1^2 + 2^2 right| = left| frac10sqrt5 right| = 2sqrt5$$d = \left| \frac{3 + 2(6) - 5}{\sqrt{1^2 + 2^2}} \right| = \left| \frac{10}{\sqrt{5}} \right| = 2\sqrt{5}$$ Calculating d^2$d^2$: d^2 = (2sqrt5)^2 = 20$$d^2 = (2\sqrt{5})^2 = 20$$ ### Pattern Recognition Short Shortcut: The maximum distance of a family of lines passing through P$P$ from the origin is simply the length OP$OP$. The line perpendicular to OP$OP$ at P$P$ is the unique farthest line. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines