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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Aicharya · Accepted Answer

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} & 	ext { and } \frac{\beta+4}{-1}=\frac{-30}{17} \Rightarrow \beta=\frac{-38}{17} \ & 	herefore \quad \beta-\alpha=\frac{-38}{17}+\frac{69}{17}=\frac{31}{17} \ & 	herefore \quad \beta-\alpha=31 / 17 \end{aligned} $

If $(\alpha, \beta)$ is the image of the point $(3,-4)$ with respect to the line $4 x-y-1=0$, then the value…

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