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Question Number 68237 by smartsmith459@gmail.com last updated on 07/Sep/19

1 . f i n d t h e e q u a t i o n o f t h e l i n e m a k i n g a n a n g l e o f 135^{°} w i t h O_{x} a n d p a s s i n g t h r o u g h t h e p o i n t s ? (- 2, 5)

2 . f i n d t h e s l o p e o f t h e l i n e t h r o u g h t h e p o i n t s (5, 3) a n d (7, 2) . f i n d (i) t h e p e r p e n d i c u l a r f o r m (i i) f i n d t h e i n t e r c e p t f o r m o f i t s e q u a t i o n .

3 . D e t e r m i n e t h e g r a d i e n t o f t h e s t r a i g h t l i n e g r a p h p a s s i n g t h r o u g h t h e c o - o r d i n a t e s :

(i) (2, 7) a n d (- 3, 4)

(i i) (\frac{1}{4}, \frac{- 3}{4}) a n d (\frac{- 1}{2}, \frac{5}{8}) .

Commented by kaivan.ahmadi last updated on 07/Sep/19

1 .

m = t a n 135 = t a n (90 + 45) = - c o t 45 = - \frac{\sqrt{2}}{2}

y - 5 = - \frac{\sqrt{2}}{2} (x + 2) \Rightarrow y = - \frac{\sqrt{2}}{2} x + (5 - \sqrt{2})

2 .

m = \frac{3 - 2}{5 - 7} = - \frac{1}{2}

y - 3 = - \frac{1}{2} (x - 5) \Rightarrow y = - \frac{1}{2} x + \frac{11}{2}

p e r p e n d i c u l a r s l o p e m_{1} = 2

3 .

(i) .

y - 7 = \frac{4 - 7}{- 3 - 2} (x - 2) \Rightarrow y = \frac{3}{5} x + \frac{29}{5} \Rightarrow

f (x, y) = 3 x - 5 y + 29 = 0

▽ f = (3, - 5)