Question

Let d be the distance between the parallel lines 3x - 2y + 5 = 0 and 3x - 2y + 5 + 2√13 = 0. Let L1 = 3x - 2y + k1 = 0 (k1 > 0) and L2 = 3x - 2y + k2 = 0 (k2 > 0) be two lines that are at the distance of \(\frac{4d}{√13}\) and \(\frac{3d}{√13}\) from the line 3x - 2y + 5y = 0. Then the combined equation of the lines L₁ = 0 and L₂ = 0 is:

• A. (3x - 2y)² + 24(3x - 2y) + 143 = 0
• B. (3x - 2y)² + 8(3x - 2y)+ 33 = 0
• C. (3x - 2y)² +12(3x-2y) + 13 = 0
• D. (3x - 2y)² +12(3x-2y) + 1 = 0

Answer 1

The Correct Option is A

To find the combined equation of two lines $L_1$ and $L_2$ that are parallel to $3x - 2y + 5 = 0$, with specific distances from it, we proceed as follows:

1. Distance Between Given Parallel Lines:
Consider the parallel lines $3x - 2y + 5 = 0$ and $3x - 2y + 5 + 2\sqrt{13} = 0$. The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by:

$ d = \frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}} $
Here, $a = 3$, $b = -2$, $c_1 = 5$, $c_2 = 5 + 2\sqrt{13}$. So:

$ d = \frac{|(5 + 2\sqrt{13}) - 5|}{\sqrt{3^2 + (-2)^2}} = \frac{2\sqrt{13}}{\sqrt{9 + 4}} = \frac{2\sqrt{13}}{\sqrt{13}} = 2 $

2. Finding $k_1$ for Line $L_1$:
Line $L_1$ is given as $3x - 2y + k_1 = 0$, and its distance from $3x - 2y + 5 = 0$ is $\frac{4d}{\sqrt{13}}$. Substituting $d = 2$:

$ \frac{4 \cdot 2}{\sqrt{13}} = \frac{8}{\sqrt{13}} $
The distance between $3x - 2y + k_1 = 0$ and $3x - 2y + 5 = 0$ is:

$ \frac{|k_1 - 5|}{\sqrt{3^2 + (-2)^2}} = \frac{|k_1 - 5|}{\sqrt{13}} $
Equating to the given distance:

$ \frac{|k_1 - 5|}{\sqrt{13}} = \frac{8}{\sqrt{13}} $
$ |k_1 - 5| = 8 $
Solving:

$ k_1 - 5 = \pm 8 $
$ k_1 = 5 + 8 = 13 \quad \text{or} \quad k_1 = 5 - 8 = -3 $
Since $k_1 > 0$, we choose:

$ k_1 = 13 $
Thus, the equation of $L_1$ is $3x - 2y + 13 = 0$.

3. Finding $k_2$ for Line $L_2$:
Line $L_2$ is given as $3x - 2y + k_2 = 0$, and its distance from $3x - 2y + 5 = 0$ is $\frac{3d}{\sqrt{13}}$. Substituting $d = 2$:

$ \frac{3 \cdot 2}{\sqrt{13}} = \frac{6}{\sqrt{13}} $
The distance between $3x - 2y + k_2 = 0$ and $3x - 2y + 5 = 0$ is:

$ \frac{|k_2 - 5|}{\sqrt{13}} $
Equating:

$ \frac{|k_2 - 5|}{\sqrt{13}} = \frac{6}{\sqrt{13}} $
$ |k_2 - 5| = 6 $
Solving:

$ k_2 - 5 = \pm 6 $
$ k_2 = 5 + 6 = 11 \quad \text{or} \quad k_2 = 5 - 6 = -1 $
Since $k_2 > 0$, we choose:

$ k_2 = 11 $
Thus, the equation of $L_2$ is $3x - 2y + 11 = 0$.

4. Forming the Combined Equation:
The combined equation of $L_1$ and $L_2$ is the product of their equations:

$ (3x - 2y + 13)(3x - 2y + 11) = 0 $
Let $u = 3x - 2y$. Then the equation becomes:

$ (u + 13)(u + 11) = 0 $
Expand:

$ u^2 + (13 + 11)u + (13 \cdot 11) = u^2 + 24u + 143 $
Substitute $u = 3x - 2y$:

$ (3x - 2y)^2 + 24(3x - 2y) + 143 = 0 $

Final Answer:
The combined equation of the lines is $ (3x - 2y)^2 + 24(3x - 2y) + 143 = 0 $.