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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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