Question

The Image of the line $3x-y=2$ in the line $y=x-1$ is

• A. $x - 3y = 2$
• B. $x + 3y = 2$
• C. $x + 3y = 5$
• D. $x - 3y = 5$

Answer 1

Answer: (A)

x - 3y = 2

Answer 2

To find the image of a line $L_1: 3x - y = 2$ in the line $L_m: y = x - 1$, we can follow these steps:

1. Find the point of intersection of the original line and the mirror line.

The intersection point $P$ will lie on both the original line and its image. Given lines: $L_1: 3x - y = 2$ $L_m: y = x - 1$

Substitute $y = x - 1$ from $L_m$ into $L_1$: $3x - (x - 1) = 2$ $3x - x + 1 = 2$ $2x + 1 = 2$ $2x = 1$ $x = \frac{1}{2}$

Now, find the corresponding $y$-coordinate using $y = x - 1$: $y = \frac{1}{2} - 1 = -\frac{1}{2}$

So, the point of intersection is $P\left(\frac{1}{2}, -\frac{1}{2}\right)$. This point must be on the image line.

2. Determine the slopes of the original line and the mirror line.

The equation of $L_1$ is $3x - y = 2$, which can be rewritten as $y = 3x - 2$. The slope of $L_1$ is $m_1 = 3$.

The equation of $L_m$ is $y = x - 1$. The slope of $L_m$ is $m_m = 1$.

3. Find the slope of the image line using the angle property.

Let $L_i$ be the image line and its slope be $m_i$. The angle between the mirror line ($L_m$) and the original line ($L_1$) is equal in magnitude but opposite in sign to the angle between the mirror line ($L_m$) and the image line ($L_i$). The tangent of the angle $\theta_1$ from $L_m$ to $L_1$ is given by: $\tan \theta_1 = \frac{m_1 - m_m}{1 + m_1 m_m} = \frac{3 - 1}{1 + (3)(1)} = \frac{2}{4} = \frac{1}{2}$

The tangent of the angle $\theta_i$ from $L_m$ to $L_i$ is given by: $\tan \theta_i = \frac{m_i - m_m}{1 + m_i m_m}$

Since $\theta_i = -\theta_1$, we have $\tan \theta_i = -\tan \theta_1$. $\frac{m_i - 1}{1 + m_i(1)} = -\frac{1}{2}$ $\frac{m_i - 1}{1 + m_i} = -\frac{1}{2}$

Cross-multiply: $2(m_i - 1) = -(1 + m_i)$ $2m_i - 2 = -1 - m_i$ $3m_i = 1$ $m_i = \frac{1}{3}$

4. Write the equation of the image line.

The image line passes through the point $P\left(\frac{1}{2}, -\frac{1}{2}\right)$ and has a slope $m_i = \frac{1}{3}$. Using the point-slope form $y - y_1 = m(x - x_1)$: $y - \left(-\frac{1}{2}\right) = \frac{1}{3}\left(x - \frac{1}{2}\right)$ $y + \frac{1}{2} = \frac{1}{3}x - \frac{1}{6}$

To eliminate fractions, multiply the entire equation by the least common multiple of the denominators (which is 6): $6\left(y + \frac{1}{2}\right) = 6\left(\frac{1}{3}x - \frac{1}{6}\right)$ $6y + 3 = 2x - 1$

Rearrange the terms to the standard form $Ax + By = C$: $2x - 6y = 3 + 1$ $2x - 6y = 4$

Divide the entire equation by 2: $x - 3y = 2$

This matches option A.