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 - y

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$-3$

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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 $\frac{h - x_1}{a} = \frac{k - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$.Given the line $5x - 3y - 2 = 0$ and the point $(2, -3)$,we have $a = 5, b = -3, c = -2, x_1 = 2, y_1 = -3$.Calculate $ax_1 + by_1 + c = 5(2) - 3(-3) - 2 = 10 + 9 - 2 = 17$.Calculate $a^2 + b^2 = 5^2 + (-3)^2 = 25 + 9 = 34$.Substitute these into the formula: $\frac{h - 2}{5} = \frac{k - (-3)}{-3} = -2 \frac{17}{34} = -2 \frac{17}{34} = -1$.So,$h - 2 = 5(-1) \implies h = -3$.And $k + 3 = -3(-1) \implies k + 3 = 3 \implies k = 0$.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 $h + k =$

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