Question

The reflection of the line 3x + 4y + 5 =0 with respect to the line 2x + y + 1 = 0 is

Answer 1

5x - 1 = 0

Answer 2

To find the reflection of a line $L_1: 3x + 4y + 5 = 0$ with respect to a mirror line $L_M: 2x + y + 1 = 0$, we can use the following approach:

1. Find the intersection point of $L_1$ and $L_M$.

   The reflected line $L_2$ must pass through the intersection point of $L_1$ and $L_M$. Given lines: $L_1: 3x + 4y + 5 = 0$ (1) $L_M: 2x + y + 1 = 0$ (2)

   From (2), $y = -2x - 1$. Substitute this into (1): $3x + 4(-2x - 1) + 5 = 0$ $3x - 8x - 4 + 5 = 0$ $-5x + 1 = 0$ $x = \frac{1}{5}$

   Substitute $x = \frac{1}{5}$ back into $y = -2x - 1$: $y = -2\left(\frac{1}{5}\right) - 1 = -\frac{2}{5} - \frac{5}{5} = -\frac{7}{5}$

   So, the intersection point $P = \left(\frac{1}{5}, -\frac{7}{5}\right)$. This point lies on the reflected line $L_2$.

2. Choose a point on $L_1$ and find its reflection with respect to $L_M$.

   Let's choose a point $Q$ on $L_1$. For instance, if $x = -3$ in $3x + 4y + 5 = 0$: $3(-3) + 4y + 5 = 0$ $-9 + 4y + 5 = 0$ $4y - 4 = 0 \Rightarrow y = 1$

   So, $Q = (-3, 1)$ is a point on $L_1$.

   Now, find the reflection of $Q(x_1, y_1) = (-3, 1)$ with respect to $L_M: Ax + By + C = 0$, where $A=2, B=1, C=1$. Let the reflected point be $Q'(x', y')$. The formula for reflection is:

   $\frac{x' - x_1}{A} = \frac{y' - y_1}{B} = -2 \frac{Ax_1 + By_1 + C}{A^2 + B^2}$

   Substitute the values: $Ax_1 + By_1 + C = 2(-3) + 1(1) + 1 = -6 + 1 + 1 = -4$ $A^2 + B^2 = 2^2 + 1^2 = 4 + 1 = 5$

   So, $\frac{x' - (-3)}{2} = \frac{y' - 1}{1} = -2 \frac{-4}{5}$ $\frac{x' + 3}{2} = \frac{y' - 1}{1} = \frac{8}{5}$

   From $\frac{x' + 3}{2} = \frac{8}{5}$: $x' + 3 = \frac{16}{5} \Rightarrow x' = \frac{16}{5} - 3 = \frac{16 - 15}{5} = \frac{1}{5}$

   From $\frac{y' - 1}{1} = \frac{8}{5}$: $y' - 1 = \frac{8}{5} \Rightarrow y' = \frac{8}{5} + 1 = \frac{8 + 5}{5} = \frac{13}{5}$

   The reflected point is $Q' = \left(\frac{1}{5}, \frac{13}{5}\right)$.

3. Find the equation of the line passing through $P$ and $Q'$.

   The reflected line $L_2$ passes through $P = \left(\frac{1}{5}, -\frac{7}{5}\right)$ and $Q' = \left(\frac{1}{5}, \frac{13}{5}\right)$. Since both points have the same x-coordinate ($x = \frac{1}{5}$), the reflected line is a vertical line. The equation of the line is $x = \frac{1}{5}$. This can be rewritten as $5x = 1$, or $5x - 1 = 0$.

Explanation of the solution:

1. Intersection Point: The intersection point of the original line ($3x+4y+5=0$) and the mirror line ($2x+y+1=0$) is found by solving the system of equations. This point, $\left(\frac{1}{5}, -\frac{7}{5}\right)$, lies on both the original line and its reflection.

2. Reflected Point: A generic point on the original line, say $(-3, 1)$, is chosen. Its reflection, $\left(\frac{1}{5}, \frac{13}{5}\right)$, across the mirror line is calculated using the standard reflection formula.

3. Equation of Reflected Line: The reflected line passes through the intersection point $\left(\frac{1}{5}, -\frac{7}{5}\right)$ and the reflected point $\left(\frac{1}{5}, \frac{13}{5}\right)$. Since both points have the same x-coordinate, the reflected line is a vertical line $x = \frac{1}{5}$, which simplifies to $5x - 1 = 0$.

Answer:

The reflection of the line $3x + 4y + 5 = 0$ with respect to the line $2x + y + 1 = 0$ is $5x - 1 = 0$.