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(A) Let $(a, b)$ be the coordinates of the foot of the perpendicular from the point $(-1, 3)$ to the line $3x - 4y - 16 = 0$.The slope of the line joi

Vedclass · Accepted Answer

$\left(\frac{68}{25}, -\frac{49}{25}\right)$

Vedclass · Answer

(A) Let $(a, b)$ be the coordinates of the foot of the perpendicular from the point $(-1, 3)$ to the line $3x - 4y - 16 = 0$.The slope of the line joining $(-1, 3)$ and $(a, b)$ is $m_1 = \frac{b - 3}{a + 1}$.The slope of the given line $3x - 4y - 16 = 0$ is $m_2 = \frac{3}{4}$.Since the two lines are perpendicular,$m_1 	imes m_2 = -1$.$	herefore \left(\frac{b - 3}{a + 1}ight) 	imes \left(\frac{3}{4}ight) = -1$$\Rightarrow 3b - 9 = -4a - 4$$\Rightarrow 4a + 3b = 5$ ..... $(1)$Since the point $(a, b)$ lies on the line $3x - 4y - 16 = 0$,we have:$3a - 4b = 16$ ..... $(2)$Solving equations $(1)$ and $(2)$:Multiply $(1)$ by $4$ and $(2)$ by $3$:$16a + 12b = 20$$9a - 12b = 48$Adding these,$25a = 68 \Rightarrow a = \frac{68}{25}$.Substituting $a$ in $(1)$: $4(\frac{68}{25}) + 3b = 5$ $\Rightarrow \frac{272}{25} + 3b = 5$ $\Rightarrow 3b = 5 - \frac{272}{25} = \frac{125 - 272}{25} = -\frac{147}{25}$ $\Rightarrow b = -\frac{49}{25}$.Thus,the coordinates of the foot of the perpendicular are $\left(\frac{68}{25}, -\frac{49}{25}ight)$.

Find the coordinates of the foot of the perpendicular from the point $(-1, 3)$ to the line $3x - 4y - 16 = 0$.

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