(x_1, y_1) is the point of concurrency of a f | vedclass

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(A) Let the equation of a line passing through the point $(x_1, y_1)$ with slope $m$ be $y - y_1 - m(x - x_1) = 0$,which can be written as $mx - y + (

Vedclass · Accepted Answer

$(1, 1)$

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(A) Let the equation of a line passing through the point $(x_1, y_1)$ with slope $m$ be $y - y_1 - m(x - x_1) = 0$,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 the lengths of the perpendiculars,we take the signed distance $\frac{mx_0 - y_0 + y_1 - mx_1}{\sqrt{m^2 + 1}}$.The sum of the perpendiculars from $(2, 0)$,$(0, 2)$,and $(1, 1)$ is:$\frac{(2m - 0 + y_1 - mx_1) + (0m - 2 + y_1 - mx_1) + (1m - 1 + y_1 - mx_1)}{\sqrt{m^2 + 1}} = 0$Simplifying the numerator:$(2m + y_1 - mx_1) + (y_1 - 2 - mx_1) + (m - 1 + y_1 - mx_1) = 0$$m(2 + 1 - 3x_1) + (3y_1 - 3) = 0$$m(3 - 3x_1) + 3(y_1 - 1) = 0$For this to hold for any slope $m$,the coefficients of $m$ and the constant term must be zero independently:$3 - 3x_1 = 0 \Rightarrow x_1 = 1$$3(y_1 - 1) = 0 \Rightarrow y_1 = 1$Thus,$(x_1, y_1) = (1, 1)$.Therefore,option $(A)$ is correct.

$(x_1, y_1)$ is the point of concurrency of a family of lines. If the algebraic sum of the lengths of the perpendiculars drawn to these lines from $(2, 0)$,$(0, 2)$,and $(1, 1)$ is zero,then $(x_1, y_1) =$

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