The coordinates of the foot of the perpendicu | vedclass

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(A) We know that the foot of the perpendicular $(h, k)$ from a point $(x_1, y_1)$ to the line $ax+by+c=0$ is given by the formula:$\frac{h-x_1}{a} = \

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$\left(\frac{9}{5}, \frac{17}{5}\right)$

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(A) We know that the foot of the perpendicular $(h, k)$ from a point $(x_1, y_1)$ to the line $ax+by+c=0$ is given by the formula:$\frac{h-x_1}{a} = \frac{k-y_1}{b} = \frac{-(ax_1+by_1+c)}{a^2+b^2}$Here,the point $(x_1, y_1) = (3,4)$ and the line is $2x+y-7=0$.Thus,$a=2, b=1, c=-7$.Substituting these values into the formula:$\frac{h-3}{2} = \frac{k-4}{1} = \frac{-(2(3) + 1(4) - 7)}{2^2 + 1^2}$$\frac{h-3}{2} = \frac{k-4}{1} = \frac{-(6+4-7)}{4+1}$$\frac{h-3}{2} = \frac{k-4}{1} = \frac{-3}{5}$Now,solving for $h$:$h-3 = 2 	imes \left(\frac{-3}{5}ight) = \frac{-6}{5}$$h = 3 - \frac{6}{5} = \frac{15-6}{5} = \frac{9}{5}$Solving for $k$:$k-4 = 1 	imes \left(\frac{-3}{5}ight) = \frac{-3}{5}$$k = 4 - \frac{3}{5} = \frac{20-3}{5} = \frac{17}{5}$Therefore,the coordinates of the foot of the perpendicular are $\left(\frac{9}{5}, \frac{17}{5}ight)$.

The coordinates of the foot of the perpendicular drawn from the point $(3,4)$ on the line $2x+y-7=0$ are

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