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 (x1 , y1 ) is y – y1 = m (x – x1 ). Taking the algebraic sum of perpendicular distances equal to zero, we get y – 1 = m (x – 1). Thus (x1 , y1 ) is (1, 1).]

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Solution

Dear student
$W e k n o w t h a t l e n g t h o f t h e p e r p e n d i c u l a r f r o m (x_{1}, y_{1}) t o t h e l i n e a x + b y + c = 0 i s p = |\frac{a x_{1} + b y_{1} + c}{\sqrt{a^{2} + b^{2}}}| L e t t h e v a r i a b l e l i n e b e a x + b y = 1 I t i s g i v e n t h a t t h e a l g e b r a i c s u m o f t h e p e r p e n d i c u l a r f r o m t h e p o i n t s (2, 0), (0, 2), (1, 1) t o t h i s l i n e i s z e r o S o \frac{(2) a + (0) b - 1}{\sqrt{a^{2} + b^{2}}} + \frac{(0) a + (2) b - 1}{\sqrt{a^{2} + b^{2}}} + \frac{a + b - 1}{\sqrt{a^{2} + b^{2}}} = 0 \Rightarrow 3 a + 3 b - 3 = 0 \Rightarrow a + b - 1 = 0 \Rightarrow a + b = 1 T h i s i s t h e l i n e a r r e l a t i o n b e t w e e n t h e p a r a m e t e r a a n d b . S o t h e e q a x + b y = 1 r e p r e s e n t s a f a m i l y o f s t r a i g h t l i n e s p a s \sin g t h r o u g h a f i x e d p o i n t C o m p a r i n g a x + b y = 1 a n d a + b = 1 w e o b t a i n t h e c o o r d i n a t e s o f t h e f i x e d p o i n t a r e (1, 1)$
Regards

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