If (h, k) is the image of the point (3, -4) 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 $\frac{h - x_1}{a} = \frac{k - y_1}{b} =

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$5$

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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 $\frac{h - x_1}{a} = \frac{k - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$.For the point $(3, -4)$ and line $2x - 3y - 5 = 0$:$\frac{h - 3}{2} = \frac{k - (-4)}{-3} = \frac{-2(2(3) - 3(-4) - 5)}{2^2 + (-3)^2} = \frac{-2(6 + 12 - 5)}{4 + 9} = \frac{-2(13)}{13} = -2$.Thus,$h - 3 = -4 \implies h = -1$ and $k + 4 = 6 \implies k = 2$.The image point is $(-1, 2)$.Now,find the foot of the perpendicular $(\ell, m)$ from $(-1, 2)$ to the line $3x + 2y + 12 = 0$ using $\frac{\ell - x_1}{a} = \frac{m - y_1}{b} = \frac{-(ax_1 + by_1 + c)}{a^2 + b^2}$:$\frac{\ell - (-1)}{3} = \frac{m - 2}{2} = \frac{-(3(-1) + 2(2) + 12)}{3^2 + 2^2} = \frac{-(-3 + 4 + 12)}{9 + 4} = \frac{-13}{13} = -1$.So,$\ell + 1 = -3 \implies \ell = -4$ and $m - 2 = -2 \implies m = 0$.The foot of the perpendicular is $(-4, 0)$.Finally,calculate $\ell h + mk + 1 = (-4)(-1) + (0)(2) + 1 = 4 + 0 + 1 = 5$.

If $(h, k)$ is the image of the point $(3, -4)$ with respect to the line $2x - 3y - 5 = 0$ and $(\ell, m)$ is the foot of the perpendicular from $(h, k)$ onto the line $3x + 2y + 12 = 0$,then $\ell h + mk + 1 =$ ?

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