If (alpha, beta) is the image of the point (3 | vedclass

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(C) Let the image of the point $(3, -4)$ with respect to the line $4x - y - 1 = 0$ be $P(\alpha, \beta)$.Using the formula for the image of a point $(

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$\frac{31}{17}$

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(C) Let the image of the point $(3, -4)$ with respect to the line $4x - y - 1 = 0$ be $P(\alpha, \beta)$.Using the formula for the image of a point $(x_1, y_1)$ with respect to the line $ax + by + c = 0$ is $\frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$.Substituting the values,we get:$\frac{\alpha - 3}{4} = \frac{\beta + 4}{-1} = \frac{-2(4(3) - (-4) - 1)}{4^2 + (-1)^2}$.$\frac{\alpha - 3}{4} = \frac{\beta + 4}{-1} = \frac{-2(12 + 4 - 1)}{16 + 1} = \frac{-2(15)}{17} = \frac{-30}{17}$.Now,solving for $\alpha$:$\frac{\alpha - 3}{4} = \frac{-30}{17}$ $\Rightarrow \alpha - 3 = \frac{-120}{17}$ $\Rightarrow \alpha = 3 - \frac{120}{17} = \frac{51 - 120}{17} = \frac{-69}{17}$.Solving for $\beta$:$\frac{\beta + 4}{-1} = \frac{-30}{17}$ $\Rightarrow \beta + 4 = \frac{30}{17}$ $\Rightarrow \beta = \frac{30}{17} - 4 = \frac{30 - 68}{17} = \frac{-38}{17}$.Finally,calculating $\beta - \alpha$:$\beta - \alpha = \frac{-38}{17} - (\frac{-69}{17}) = \frac{-38 + 69}{17} = \frac{31}{17}$.

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

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