Key Concepts and Formulas

• Family of Lines Passing Through a Fixed Point: The equation $L_1 + \lambda L_2 = 0$ represents a family of lines passing through the intersection of lines $L_1 = 0$ and $L_2 = 0$.
• Distance of a Point from a Line: The distance $D$ of a point $(x_1, y_1)$ from a line $Ax + By + C = 0$ is given by $D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
• Line Farthest from the Origin: For a family of lines passing through a fixed point $P$, the line farthest from the origin is perpendicular to the line segment $OP$, where $O$ is the origin.

Step-by-Step Solution

Step 1: Finding the Fixed Point P

We need to find the point P through which all lines in the given family pass.

Why this step? Expressing the given equation in the form $L_1 + \lambda L_2 = 0$ allows us to find the intersection of $L_1 = 0$ and $L_2 = 0$, which gives us the fixed point P.

The equation of the family of lines is: $x(3\lambda + 1) + y(7\lambda + 2) = 17\lambda + 5$ Expanding and rearranging: $3\lambda x + x + 7\lambda y + 2y = 17\lambda + 5$ Grouping terms with and without $\lambda$: $(x + 2y - 5) + (3\lambda x + 7\lambda y - 17\lambda) = 0$ Factoring out $\lambda$: $(x + 2y - 5) + \lambda(3x + 7y - 17) = 0$ This is in the form $L_1 + \lambda L_2 = 0$, where: $L_1: x + 2y - 5 = 0$ $L_2: 3x + 7y - 17 = 0$

Solving the system of linear equations:

1. $x + 2y = 5$
2. $3x + 7y = 17$

From equation (1), $x = 5 - 2y$. Substituting into equation (2): $3(5 - 2y) + 7y = 17$ $15 - 6y + 7y = 17$ $y = 2$

Substituting $y = 2$ back into $x = 5 - 2y$: $x = 5 - 2(2) = 1$ So, the fixed point P is $(1, 2)$.

Step 2: Determining the Line L Farthest from the Origin

We need to find the line $L$ that is farthest from the origin.

Why this step? We use the geometric property that the line $L$ farthest from the origin is perpendicular to the line segment OP.

The origin is O $(0,0)$ and the fixed point is P $(1,2)$. The slope of OP ($m_{OP}$) is $\frac{2 - 0}{1 - 0} = 2$. Since line $L$ is perpendicular to OP, its slope ($m_L$) is $-\frac{1}{m_{OP}} = -\frac{1}{2}$.

Using the point-slope form: $y - y_1 = m(x - x_1)$ with point $(1,2)$ and slope $-\frac{1}{2}$: $y - 2 = -\frac{1}{2}(x - 1)$ $2(y - 2) = -(x - 1)$ $2y - 4 = -x + 1$ $x + 2y - 5 = 0$ So, the equation of line $L$ is $x + 2y - 5 = 0$.

Step 3: Calculating the Distance $d$

We need to find the distance of line $L: x + 2y - 5 = 0$ from the point $(3,6)$.

Why this step? This is a direct application of the distance formula.

Here, $(x_1, y_1) = (3,6)$ and the line is $x + 2y - 5 = 0$, so $A=1$, $B=2$, $C=-5$. Using the distance formula: $d = \frac{|(1)(3) + (2)(6) - 5|}{\sqrt{1^2 + 2^2}} = \frac{|3 + 12 - 5|}{\sqrt{5}} = \frac{10}{\sqrt{5}}$

Step 4: Final Calculation: $d^2$

We need to find the value of $d^2$. $d^2 = \left(\frac{10}{\sqrt{5}}\right)^2 = \frac{100}{5} = 20$

Common Mistakes & Tips

• Carefully rearrange the equation to isolate the parameter $\lambda$ to correctly identify $L_1$ and $L_2$.
• Remember that the line farthest from the origin is perpendicular to the line segment connecting the origin and the fixed point.
• Double-check the arithmetic in the distance formula calculation.

Summary

We found the fixed point P by rearranging the equation into the form $L_1 + \lambda L_2 = 0$. Then, we used the geometric property that the line farthest from the origin is perpendicular to OP to find the equation of line L. Finally, we calculated the distance $d$ of the point $(3,6)$ from line L and computed $d^2$.

The final answer is \boxed{20}, which corresponds to option (B).