Question

Consider the lines $ x(3\lambda + 1) + y(7\lambda + 2) = 17\lambda + 5 $. If P is the point through which all these lines pass and the distance of L from the point $ Q(3, 6) $ is \( d \), then the distance of L from the point \( (3, 6) \) is \( d \), then the value of \( d^2 \) is

• A. 20
• B. 30
• C. 10
• D. 15

Hint:

A family of lines of the form \( a_1\lambda + b_1 + (a_2\lambda + b_2) = 0 \) passes through the intersection of the lines \( a_1 = 0 \) and \( b_1 = 0 \). Rearrange the given equation to find the two lines whose intersection point P lies on all the lines of the family. Then, use the distance formula to find the distance between point P and the given point Q, and finally square this distance.

Answer 1

The Correct Option is A

The given equation of the family of lines is: \[ x(3\lambda + 1) + y(7\lambda + 2) = 17\lambda + 5 \] Rearranging the terms to group by \( \lambda \): \[ 3\lambda x + x + 7\lambda y + 2y = 17\lambda + 5 \] \[ \lambda (3x + 7y - 17) + (x + 2y - 5) = 0 \] This represents a family of lines passing through the intersection of the lines: \[ 3x + 7y - 17 = 0 \quad \text{(i)} \] \[ x + 2y - 5 = 0 \quad \text{(ii)} \] Multiply equation (ii) by 3: \[ 3x + 6y - 15 = 0 \quad \text{(iii)} \] Subtract (iii) from (i): \[ (3x + 7y - 17) - (3x + 6y - 15) = 0 \] \[ y - 2 = 0 \Rightarrow y = 2 \] Substitute \( y = 2 \) into equation (ii): \[ x + 2(2) - 5 = 0 \] \[ x + 4 - 5 = 0 \Rightarrow x = 1 \] So, the point \( P \) through which all lines pass is: \[ P = (1, 2) \] The distance from \( P(1, 2) \) to point \( Q(3, 6) \) is: \[ d = \sqrt{(3 - 1)^2 + (6 - 2)^2} \] \[ = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} \] Therefore, \[ d^2 = (\sqrt{20})^2 = 20 \]

Answer 2

To solve this problem, we need to find the point \( P \) through which the line \( x(3\lambda + 1) + y(7\lambda + 2) = 17\lambda + 5 \) passes for all values of \(\lambda\). After that, we calculate the distance of this line from the point \( Q(3, 6) \) to eventually determine \( d^2 \).

First, observe the line equation:

\(x(3\lambda + 1) + y(7\lambda + 2) = 17\lambda + 5\)

Rearranging, it becomes:

\(x(3\lambda + 1) + y(7\lambda + 2) - 17\lambda - 5 = 0\)

We need a point \( P(h, k) \) such that it satisfies the equation for any \(\lambda\). Let us make the coefficients of both \( x \) and \( y \) equal to zero for their \(\lambda\) components to find a relationship:

1. Coefficient of \(\lambda\):
   • \(3h + 7k - 17 = 0\)
2. Constant term:
   • \(h + 2k - 5 = 0\)

Now, solving these two equations simultaneously:

\(3h + 7k = 17\)   (Equation 1)

\(h + 2k = 5\)   (Equation 2)

Let's multiply Equation 2 by 3:

• \(3h + 6k = 15\)

Subtract this new equation from Equation 1:

• \((3h + 7k) - (3h + 6k) = 17 - 15\)
• \(k = 2\)

Substitute \( k = 2 \) back into Equation 2:

• \(h + 2(2) = 5\)
• \(h + 4 = 5\)
• \(h = 1\)

Therefore, the point \( P(1, 2) \) lies on all lines. Now compute the distance from point \( Q(3, 6) \) to the line through \( P \). The standard form of the line through \( P \) can be rewritten as:

\(3(x - 1) + 7(y - 2) = 0\)

This simplifies to:

\(3x + 7y = 17\)

The distance \( d \) from point \( Q(3, 6) \) to this line is calculated using the distance formula:

\(d = \frac{|3(3) + 7(6) - 17|}{\sqrt{3^2 + 7^2}}\)

Calculate the numerator:

• \(|9 + 42 - 17| = |34| = 34\)

Calculate the denominator:

• \(\sqrt{9 + 49} = \sqrt{58}\)

Thus, the distance \( d \) is:

\(d = \frac{34}{\sqrt{58}}\)

Finally, the value of \( d^2 \) is:

\(d^2 = \left(\frac{34}{\sqrt{58}}\right)^2 = \frac{1156}{58} = 20\)

Thus, the value of \( d^2 \) is 20. Therefore, the correct answer is:

1. Option: 20