Question-20744

Question Number 20744 by mondodotto@gmail.com last updated on 02/Sep/17

Answered by $@ty@m last updated on 02/Sep/17

A T Q,

L e t t h e e q u a t i o n o f t h e s t . l i n e b e

y = m x + c - - (1)

I t p a s s e s t h r o u g h (4, - 2)

∴ - 2 = 4 m + c

\Rightarrow c = - 2 - 4 m

∴ t h e e q u a t i o n o f t h e s t . l i n e i s

y = m x - 4 m - 2 - - (2)

P u t y = 0

\Rightarrow x = \frac{4 m + 2}{m}

∴ c o o r d i n a t e s o f R = (\frac{4 m + 2}{m}, 0)

P u t x = 0

\Rightarrow y = - 4 m - 2

∴ c o o r d i n a t e s o f S = (0, - 4 m - 2)

T i s m i d p o i n t o f R S

∴ c o o r d i n a t e s o f T = (\frac{1 + 2 m}{2}, - 2 m - 1) = (p, q), s a y

\Rightarrow p = \frac{1 + 2 m}{2}, q = - 2 m - 1

\Rightarrow p = \frac{- q}{2}

\Rightarrow 2 p + q = 0

∴ l o c u s o f T (p, q) i s

2 x + y = 0

w h i c h s a t i s f i e s (0, 0)

\Rightarrow i t p a s s e s t h r o u g h o r i g i n .

Commented by mondodotto@gmail.com last updated on 02/Sep/17

thanx a lot!

Commented by mondodotto@gmail.com last updated on 02/Sep/17

please recheck!

Commented by $@ty@m last updated on 02/Sep/17

I s t h i s c o r r e c t n o w ?

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