The coordinates of the foot of the perpendicu | vedclass

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(B) The given line is $3x - y + 4 = 0$. Let the foot of the perpendicular from $P(2, 3)$ be $Q(h, k)$. The slope of the line is $m_1 = 3$. The slope o

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$\left( - \frac{1}{10}, \frac{37}{10} \right)$

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(B) The given line is $3x - y + 4 = 0$. Let the foot of the perpendicular from $P(2, 3)$ be $Q(h, k)$. The slope of the line is $m_1 = 3$. The slope of the perpendicular line $PQ$ is $m_2 = -\frac{1}{3}$. The equation of line $PQ$ passing through $(2, 3)$ is $y - 3 = -\frac{1}{3}(x - 2)$,which simplifies to $3y - 9 = -x + 2$,or $x + 3y = 11$. Solving the system of equations $3x - y = -4$ and $x + 3y = 11$: Multiply the first equation by $3$: $9x - 3y = -12$. Adding this to $x + 3y = 11$ gives $10x = -1$,so $x = -\frac{1}{10}$. Substituting $x$ into $x + 3y = 11$: $-\frac{1}{10} + 3y = 11 \implies 3y = 11 + \frac{1}{10} = \frac{111}{10} \implies y = \frac{37}{10}$. Thus,the foot of the perpendicular is $\left( - \frac{1}{10}, \frac{37}{10} \right)$.

The coordinates of the foot of the perpendicular from the point $(2, 3)$ on the line $y = 3x + 4$ are given by

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