If (h, k) is the image of the point (2, -3) w | vedclass

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(A) The formula for the image $(h, k)$ of a point $(x_1, y_1)$ with respect to the line $ax + by + c = 0$ is given by: $\frac{h - x_1}{a} = \frac{k -

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(A) The formula for the image $(h, k)$ of a point $(x_1, y_1)$ with respect to the line $ax + by + c = 0$ is given by: $\frac{h - x_1}{a} = \frac{k - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$ Here,$(x_1, y_1) = (2, -3)$ and the line is $5x - 3y - 2 = 0$. Substituting the values: $\frac{h - 2}{5} = \frac{k - (-3)}{-3} = \frac{-2(5(2) - 3(-3) - 2)}{5^2 + (-3)^2}$ $\frac{h - 2}{5} = \frac{k + 3}{-3} = \frac{-2(10 + 9 - 2)}{25 + 9}$ $\frac{h - 2}{5} = \frac{k + 3}{-3} = \frac{-2(17)}{34} = \frac{-34}{34} = -1$ Now,solving for $h$ and $k$: $\frac{h - 2}{5} = -1$ $\Rightarrow h - 2 = -5$ $\Rightarrow h = -3$ $\frac{k + 3}{-3} = -1$ $\Rightarrow k + 3 = 3$ $\Rightarrow k = 0$ Therefore,$h + k = -3 + 0 = -3$.

If $(h, k)$ is the image of the point $(2, -3)$ with respect to the line $5x - 3y = 2$,then $h + k$ is equal to

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