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Question

(B) Let the fixed point $P$ be $(x_1, y_1)$. The equation of a line passing through $P$ with slope $m$ is $y - y_1 = m(x - x_1)$,which can be written

Vedclass · Accepted Answer

$(1, 1)$

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(B) Let the fixed point $P$ be $(x_1, y_1)$. The equation of a line passing through $P$ with slope $m$ is $y - y_1 = m(x - x_1)$,which can be written as $mx - y + (y_1 - mx_1) = 0$.The perpendicular distance from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is given by $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$. Since we are considering the algebraic sum of perpendiculars,we use the signed distance $\frac{Ax_0 + By_0 + C}{\sqrt{A^2 + B^2}}$.The sum of the perpendiculars from $(2, 0)$,$(0, 2)$,and $(1, 1)$ is:$\frac{2m - 0 + y_1 - mx_1}{\sqrt{m^2 + 1}} + \frac{0 - 2 + y_1 - mx_1}{\sqrt{m^2 + 1}} + \frac{m - 1 + y_1 - mx_1}{\sqrt{m^2 + 1}} = 0$Simplifying the numerator:$(2m + y_1 - mx_1) + (y_1 - mx_1 - 2) + (m - 1 + y_1 - mx_1) = 0$$3m - 3mx_1 + 3y_1 - 3 = 0$$3(m(1 - x_1) + (y_1 - 1)) = 0$Since the line is variable,this equation must hold for all values of $m$. Thus,the coefficients of $m$ and the constant term must be zero independently:$1 - x_1 = 0 \Rightarrow x_1 = 1$$y_1 - 1 = 0 \Rightarrow y_1 = 1$Therefore,the coordinates of $P$ are $(1, 1)$.

$A$ variable line passes through a fixed point $P$. If the algebraic sum of the perpendiculars drawn from $(2, 0)$,$(0, 2)$,and $(1, 1)$ to the line is zero,then the coordinates of the point $P$ are:

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