The coordinates of the foot of the perpendicu | vedclass

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(D) Let the foot of the perpendicular be $(h, k)$.Since $(h, k)$ lies on the line $3x - y - 1 = 0$,we have $3h - k - 1 = 0 \implies k = 3h - 1$ ... $(

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$(1, 2)$

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(D) Let the foot of the perpendicular be $(h, k)$.Since $(h, k)$ lies on the line $3x - y - 1 = 0$,we have $3h - k - 1 = 0 \implies k = 3h - 1$ ... $(i)$.The slope of the given line $3x - y - 1 = 0$ is $m_1 = 3$.The slope of the line segment joining $(-2, 3)$ and $(h, k)$ is $m_2 = \frac{k - 3}{h - (-2)} = \frac{k - 3}{h + 2}$.Since the lines are perpendicular,$m_1 	imes m_2 = -1$,so $3 	imes \frac{k - 3}{h + 2} = -1$.$3(k - 3) = -(h + 2) \implies 3k - 9 = -h - 2 \implies h + 3k = 7$ ... $(ii)$.Substituting $k = 3h - 1$ from $(i)$ into $(ii)$:$h + 3(3h - 1) = 7 \implies h + 9h - 3 = 7 \implies 10h = 10 \implies h = 1$.Substituting $h = 1$ into $(i)$,$k = 3(1) - 1 = 2$.Thus,the coordinates of the foot of the perpendicular are $(1, 2)$.

The coordinates of the foot of the perpendicular drawn from the point $(-2, 3)$ to the line $3x - y - 1 = 0$ are:

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