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(A) Let the foot of the perpendicular from the origin $(0, 0)$ to the line $3x + 4y - 5 = 0$ be $(h, k)$.The slope of the given line $3x + 4y - 5 = 0$

Vedclass · Accepted Answer

$\left( \frac{3}{5}, \frac{4}{5} \right)$

Vedclass · Answer

(A) Let the foot of the perpendicular from the origin $(0, 0)$ to the line $3x + 4y - 5 = 0$ be $(h, k)$.The slope of the given line $3x + 4y - 5 = 0$ is $m_1 = -\frac{3}{4}$.The slope of the perpendicular line passing through the origin $(0, 0)$ and $(h, k)$ is $m_2 = \frac{k - 0}{h - 0} = \frac{k}{h}$.Since the lines are perpendicular,$m_1 	imes m_2 = -1$,so $\left( -\frac{3}{4} ight) 	imes \left( \frac{k}{h} ight) = -1$,which implies $\frac{k}{h} = \frac{4}{3}$,or $3k = 4h \implies k = \frac{4h}{3}$.Since the point $(h, k)$ lies on the line $3x + 4y - 5 = 0$,we have $3h + 4k = 5$.Substituting $k = \frac{4h}{3}$ into the equation: $3h + 4\left( \frac{4h}{3} ight) = 5$.$3h + \frac{16h}{3} = 5 \implies \frac{9h + 16h}{3} = 5 \implies 25h = 15 \implies h = \frac{15}{25} = \frac{3}{5}$.Now,$k = \frac{4}{3} 	imes \frac{3}{5} = \frac{4}{5}$.Thus,the foot of the perpendicular is $\left( \frac{3}{5}, \frac{4}{5} ight)$.

The foot of the perpendicular drawn from the origin to the line $3x + 4y - 5 = 0$ is

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