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, \[ d^2 = (\sqrt{20})^2 = 20 \]