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.

(Hint: Let the slope of the line be m. Then the equation of the line passing through the fixed-point P (x?, y?) is y – y ? = m (x – x?). Taking the algebraic sum of perpendicular distances equal to zero, we get y – 1 = m (x – 1). Thus (x?, y?) is (1, 1).

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Vishal Baghel | Contributor-Level 10

3 months ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t (x_{1}, y_{1}) b e t h e c o o r d i n a t e s o f t h e g i v e n P a n d m b e t h e s l o p e o f t h e l i n e . \\ ∴ E q u a t i o n o f t h e l i n e i s y - y_{1} = m (x - x_{1}) \dots (i) \\ G i v e n a r e A (2, 0), B (0, 2) a n d C (1, 1) . \\ P e r p e n d i c u l a r f r o m A (2, 0) d_{1} (s a y) \\ d_{1} = \frac{0 - y_{1} - m (2 - x_{1})}{\sqrt[]{1 + m^{2}}} \\ P e r p e n d i c u l a r f r o m B (0, 2) d_{2} (s a y) \\ d_{2} = \frac{2 - y_{1} - m (0 - x_{1})}{\sqrt[]{1 + m^{2}}} \\ S i m i l a r l y, p e r p e n d i c u l a r f r o m C (1, 1) d_{3} (s a y) \\ d_{3} = \frac{1 - y_{1} - m (1 - x_{1})}{\sqrt[]{1 + m^{2}}} \\ A c c o r d i n g t o t h e q u e s t i o n, w e h a v e \\ d_{1} + d_{2} + d_{3} = 0 \\ ∴ \frac{0 - y_{1} - m (2 - x_{1})}{\sqrt[]{1 + m^{2}}} + \frac{2 - y_{1} - m (0 - x_{1})}{\sqrt[]{1 + m^{2}}} + \frac{1 - y_{1} - m (1 - x_{1})}{\sqrt[]{1 + m^{2}}} = 0 \\ \Rightarrow - y_{1} - 2 m + m x_{1} + 2 - y_{1} + m x_{1} + 1 - y_{1} - m + m x_{1} = 0 \\ \Rightarrow 3 m x_{1} - 3 y_{1} - 3 m + 3 = 0 \Rightarrow m x_{1} - y_{1} - m + 1 = 0 \\ t h e (1, 1) s a t i f i e s t h e a b o v e e q u a t i o n . \\ H e n c e, t h e (1, 1) l i e s o n t h e l i n e . \end{array}$

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