Straight Line Problem - Detailed Solution and Interactive Quiz

Straight Line Problem [IIT 1963]

Question: The equations of the lines through the point of intersection of the lines \( x - y + 1 = 0 \) and \( 2x - 3y + 5 = 0 \) and whose distance from the point \( (3, 2) \) is \( \frac{7}{5} \) are:

Select the Correct Option

(a) \( 3x - 4y - 6 = 0 \) and \( 4x + 3y + 1 = 0 \)

(b) \( 3x - 4y + 6 = 0 \) and \( 4x - 3y - 1 = 0 \)

(c) \( 3x - 4y + 6 = 0 \) and \( 4x - 3y + 1 = 0 \)

(d) None of these

Detailed Step-by-Step Explanation

Step 1: The general equation of a line passing through the intersection of \( x - y + 1 = 0 \) and \( 2x - 3y + 5 = 0 \) can be written as:

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

Expanding and collecting like terms:

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

Here, \( \lambda \) is a parameter.

Step 2: The distance \( d \) from a point \( (x_0, y_0) \) to a line \( Ax + By + C = 0 \) is given by:

\( d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \)

For the point \( (3, 2) \) and the line \( (1+2\lambda)x + (-1-3\lambda)y + (1+5\lambda) = 0 \), we have:

\( d = \frac{|(1+2\lambda) \cdot 3 + (-1-3\lambda) \cdot 2 + (1+5\lambda)|}{\sqrt{(1+2\lambda)^2 + (-1-3\lambda)^2}} \)

Step 3: Simplify the numerator:

\( 3(1+2\lambda) - 2(1+3\lambda) + (1+5\lambda) \)

Calculate each term:
\( 3 + 6\lambda - 2 - 6\lambda + 1 + 5\lambda = 2 + 5\lambda \)

Step 4: The denominator is:

\( \sqrt{(1+2\lambda)^2 + (-1-3\lambda)^2} \)

Expanding both squares:
\( (1+2\lambda)^2 = 1 + 4\lambda + 4\lambda^2 \)
\( (-1-3\lambda)^2 = 1 + 6\lambda + 9\lambda^2 \)
Sum: \( 2 + 10\lambda + 13\lambda^2 \)

Step 5: We are given \( d = \frac{7}{5} \). Setting up the equation:

\( \frac{|2+5\lambda|}{\sqrt{2+10\lambda+13\lambda^2}} = \frac{7}{5} \)

Squaring both sides gives:

\( \frac{(2+5\lambda)^2}{2+10\lambda+13\lambda^2} = \frac{49}{25} \)

Cross-multiplying:

\( 25(2+5\lambda)^2 = 49(2+10\lambda+13\lambda^2) \)

Expanding:
LHS: \( 25(4+20\lambda+25\lambda^2) = 100 + 500\lambda + 625\lambda^2 \)
RHS: \( 49(2+10\lambda+13\lambda^2) = 98 + 490\lambda + 637\lambda^2 \)

Step 6: Equate and simplify:

\( 100 + 500\lambda + 625\lambda^2 = 98 + 490\lambda + 637\lambda^2 \)

Rearranging, we get:
\( 2 + 10\lambda - 12\lambda^2 = 0 \)
Dividing by 2:
\( 1 + 5\lambda - 6\lambda^2 = 0 \)
Rearranging:
\( 6\lambda^2 - 5\lambda - 1 = 0 \)

Step 7: Solve the quadratic:
\( \lambda = \frac{5 \pm \sqrt{25 + 24}}{12} = \frac{5 \pm \sqrt{49}}{12} = \frac{5 \pm 7}{12} \)
Thus, \( \lambda = 1 \) or \( \lambda = -\frac{1}{6} \).

Step 8: Substitute back into the line equation.

For \( \lambda = 1 \):
Coefficients become:
\( A = 1 + 2(1) = 3 \), \( B = -1 - 3(1) = -4 \), \( C = 1 + 5(1) = 6 \)
So the line is:
\( 3x - 4y + 6 = 0 \)

For \( \lambda = -\frac{1}{6} \):
\( A = 1 + 2\left(-\frac{1}{6}\right) = 1 - \frac{1}{3} = \frac{2}{3} \)
\( B = -1 - 3\left(-\frac{1}{6}\right) = -1 + \frac{1}{2} = -\frac{1}{2} \)
\( C = 1 + 5\left(-\frac{1}{6}\right) = 1 - \frac{5}{6} = \frac{1}{6} \)
Multiplying by 6 to clear fractions gives:
\( 4x - 3y + 1 = 0 \)

Final Answer: The required lines are:
\( 3x - 4y + 6 = 0 \) and \( 4x - 3y + 1 = 0 \)

This corresponds to option (c).