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If $(\alpha, \beta)$ is the image of the point $(3,-4)$ with respect to the line $4 x-y-1=0$, then the value of $\beta-\alpha$ is equal to
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Verified Answer
The correct answer is:
$\frac{31}{17}$
Let the image be $P(\alpha, \beta)$ of $(3,-4)$ with respect to line $4 x-y-1=0$.
Thus,
$$
\begin{aligned}
& \frac{\alpha-3}{4}=\frac{\beta+4}{-1}=\frac{-2(12+4-1)}{4^2+(-1)^2}=-\frac{30}{17} \\
\Rightarrow & \frac{\alpha-3}{4}=\frac{-30}{17} \Rightarrow \alpha=\frac{-69}{17}
\end{aligned}
$$
$$
\begin{aligned}
& \text { and } \frac{\beta+4}{-1}=\frac{-30}{17} \Rightarrow \beta=\frac{-38}{17} \\
& \therefore \quad \beta-\alpha=\frac{-38}{17}+\frac{69}{17}=\frac{31}{17} \\
& \therefore \quad \beta-\alpha=31 / 17
\end{aligned}
$$
Thus,
$$
\begin{aligned}
& \frac{\alpha-3}{4}=\frac{\beta+4}{-1}=\frac{-2(12+4-1)}{4^2+(-1)^2}=-\frac{30}{17} \\
\Rightarrow & \frac{\alpha-3}{4}=\frac{-30}{17} \Rightarrow \alpha=\frac{-69}{17}
\end{aligned}
$$
$$
\begin{aligned}
& \text { and } \frac{\beta+4}{-1}=\frac{-30}{17} \Rightarrow \beta=\frac{-38}{17} \\
& \therefore \quad \beta-\alpha=\frac{-38}{17}+\frac{69}{17}=\frac{31}{17} \\
& \therefore \quad \beta-\alpha=31 / 17
\end{aligned}
$$
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