a-square-s-one-diagonal-s-terminal-points-are-5-0-9-4-what-are-the-coordinate-of-other-diagonal-s-terminal-points-

Question Number 121393 by faysal last updated on 07/Nov/20

a {s q u a r e}^{'} s o n e {d i a g o n a l}^{'} s t e r m i n a l p o i n t s a r e

(5, 0), (9, 4) w h a t a r e t h e c o o r d i n a t e o f

o t h e r {d i a g o n a l}^{'} s t e r m i n a l p o i n t s

Answered by TANMAY PANACEA last updated on 07/Nov/20

D i s g o n a l e q n \frac{y - 0}{0 - 4} = \frac{x - 5}{5 - 9} \to x - y - 5 = 0

m i d p o i n t o f (5, 0) a n d (9, 4) = (7, 2)

a n o t h e r d i a g o n a l i s ⊥ (x - y - 5 = 0) i s

x + y = k w h i c h p a s s e s t h r o u g h (7, 2) \to 7 + 2 = k

E q n a n o t h d r d i a g o n a l x + y = 9

t h e s i d e o f s q u a r e m a k e 45^{o} w i t h d i a g o n a l

x - y = 5 s n d w h i c h p a s s e s t h r o u g h (5, 0) i s

y = m (x - 5) \to s l o p e o f x - y = 5 i s 1

t a n 45^{0} = \frac{1 - m}{1 + m} \to m = 0

e q n o f s i d e i s y = 0 (x a x i s)

s o l v e y = 0 a n d x + y = 9

(9, 0) i s o n e t e r m i n a l o f a n o t h e r d i a g o n a l

t h e s i d e o f s q u a r e m a k e 45^{o} w i t h d i a g o n a l x - y = 5

a n d p a s s e s t h r o u g h (9, 4) i s (y - 4) = m_{1} (x - 9)

t a n 45^{o} = \frac{1 - m_{1}}{1 + m_{1}} \to m_{1} = 0 e q n o f a n o t h d r s i d e y = 4

s o l v e y = 4 a n d x + y = 9

a n o t h e r t e r m i n a l (5, 4)

R e w u i r e d a n s w e r i s (9, 0) a n d (5, 4)

Answered by mr W last updated on 07/Nov/20

\frac{5 + 9}{2} = 7

\frac{0 + 4}{2} = 2

\frac{4 - 0}{9 - 5} = 1

\sqrt{{(9 - 5)}^{2} + {(4 - 0)}^{2}} = 4 \sqrt{2}

(7, 2) \pm \frac{4 \sqrt{2}}{2} (\frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}}) = (7 \pm 2, 2 \mp 2)

= (9, 0) a n d (5, 4)

Facebook Tweet Pin