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Question
A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2, 0), (0, 2) and (1, 1) on the line is zero. Find the coordinates of the point P.
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Solution
Let (x1, y1) be the coordinates of the given point P and m be the slope of the line.
∴ Equation of the line is y – y1 = m(x – x1) .......(i)
Given points are A(2, 0), B(0, 2) and C(1, 1).
Perpendicular distance from A(2, 0) to the line (i) d1 (say)
d1 = `(0 - y_1 - m(2 - x_1))/sqrt(1 + m^2)`
Perpendicular distance from B(0, 2) d2 (say)
d2 = `(2 - y_1 - m(0 - x_1))/sqrt(1 + m^2)`
Similarly, perpendicular distance from C(1, 1) d3 (say)
d3 = `(1 - y_1 - m(1 - x_1))/sqrt(1 + m^2)`
We have d1 + d2 + d3 = 0
∴ `(0 - y_1 - m(2 - x_1))/sqrt(1 + m^2) + (2 - y_1 - m(0 - x_1))/sqrt(1 + m^2) + (1 - y_1 - m(1 - x_1))/sqrt(1 + m^2)` = 0
⇒ – y1 – 2m + mx1 + 2 – y1 + mx1 + 1 – y1 – m + mx1 = 0
⇒ 3mx1 – 3y1 – 3m + 3 = 0
⇒ mx1 – y1 – m + 1 = 0
Since the point (1, 1) satisfies the above equation.
Hence, the point (1, 1) lies on the line.
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