Find the equation of the two lines throught (2,−3) which makes 45° with the line 2x − y = 2..hence find the cosine of the acute between the lines l_1 : y− 2x + 5=0 and l_2 : y − x + 6 (leave your answer in surd form)


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Question Number 38049 by Rio Mike last updated on 21/Jun/18

$F i n d t h e e q u a t i o n o f t h e t w o l i n e s$ $t h r o u g h t (2, - 3) w h i c h m a k e s 45 °$ $w i t h t h e l i n e 2 x - y = 2 . . h e n c e$ $f i n d t h e c o s i n e o f t h e a c u t e b e t w e e n$ $t h e l i n e s l_{1} : y - 2 x + 5 = 0 a n d$ $l_{2} : y - x + 6 (l e a v e y o u r a n s w e r i n$ $s u r d f o r m)$
	Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jun/18

	$t h e s l o p e o f s t l i n e 2 x - y = 2 i s 2$ $t h e e q n o f r e q u i r e d s t l i n e s p a s s i n g t h r o u g b$ $(2, - 3) i s y + 3 = m (x - 2)$ $g i v e n t a n 45^{o} = \frac{m - 2}{1 + 2 m} = 1$ $1 + 2 m = m - 2$ $m = - 3$ $o r t a n 45^{o} = \frac{2 - m}{1 + 2 m} = 1$ $1 + 2 m = 2 - m$ $3 m = 1$ $m = \frac{1}{3}$ $r e q u i r e d t w o l i n e s a r e y + 3 = (- 3) (x - 2)$ $y + 3 = - 3 x + 6$ $y + 3 x = 3$ $a n o t h e r e q n$ $y + 3 = \frac{1}{3} (x - 2)$ $3 y + 9 = x - 2$ $3 y - x + 11 = 0$ $s l o p e o f y - 2 x + 5 = 0 i s 2$ $s l o p e o f y - x + 5 = 0 i s 1$ $s o t a n θ = \frac{2 - 1}{1 + 2 \times 1} = \frac{1}{3}$ $θ = {t a n}^{- 1} (\frac{1}{3})$ $a c u t e a n g l e c o s θ = \frac{3}{\sqrt{3^{2} + 1^{2}}} = \frac{3}{\sqrt{10}}$
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