The coordinates of the foot of the perpendicu | vedclass

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(B) Let the given point be $P(2, 3)$ and the line be $L: x + y - 11 = 0$.The slope of the line $L$ is $m_1 = -1$.The slope of the line perpendicular t

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$(5, 6)$

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(B) Let the given point be $P(2, 3)$ and the line be $L: x + y - 11 = 0$.The slope of the line $L$ is $m_1 = -1$.The slope of the line perpendicular to $L$ is $m_2 = -\frac{1}{m_1} = 1$.The equation of the line passing through $(2, 3)$ with slope $1$ is $y - 3 = 1(x - 2)$,which simplifies to $x - y + 1 = 0$.The foot of the perpendicular is the intersection of $x + y = 11$ and $x - y = -1$.Adding the two equations: $(x + y) + (x - y) = 11 - 1 \implies 2x = 10 \implies x = 5$.Substituting $x = 5$ into $x + y = 11$: $5 + y = 11 \implies y = 6$.Thus,the foot of the perpendicular is $(5, 6)$.

The coordinates of the foot of the perpendicular from the point $(2, 3)$ on the line $x + y - 11 = 0$ are

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