Find the coordinates of the foot of the perpe | vedclass

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(A) Let the foot of the perpendicular be $P(h, k)$.Given the line $2x + 3y - 4 = 0$,the slope of this line is $m_1 = -\frac{2}{3}$.The slope of the pe

Vedclass · Accepted Answer

$\left( \frac{23}{13}, \frac{2}{13} \right)$

Vedclass · Answer

(A) Let the foot of the perpendicular be $P(h, k)$.Given the line $2x + 3y - 4 = 0$,the slope of this line is $m_1 = -\frac{2}{3}$.The slope of the perpendicular line passing through $(7, 8)$ and $(h, k)$ is $m_2 = \frac{k - 8}{h - 7}$.Since the lines are perpendicular,$m_1 	imes m_2 = -1$,so $(-\frac{2}{3}) 	imes (\frac{k - 8}{h - 7}) = -1$,which simplifies to $2(k - 8) = 3(h - 7)$,or $3h - 2k = 5$.Since $P(h, k)$ lies on the line $2x + 3y - 4 = 0$,we have $2h + 3k = 4$.Solving the system of equations:$3h - 2k = 5$ (multiply by $3$): $9h - 6k = 15$$2h + 3k = 4$ (multiply by $2$): $4h + 6k = 8$Adding these gives $13h = 23$,so $h = \frac{23}{13}$.Substituting $h$ into $2h + 3k = 4$: $2(\frac{23}{13}) + 3k = 4 \implies \frac{46}{13} + 3k = 4 \implies 3k = 4 - \frac{46}{13} = \frac{52 - 46}{13} = \frac{6}{13} \implies k = \frac{2}{13}$.The foot of the perpendicular is $\left( \frac{23}{13}, \frac{2}{13} ight)$.

Find the coordinates of the foot of the perpendicular drawn from the point $(7, 8)$ to the line $2x + 3y - 4 = 0$.

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