orthocentre of a triangle whose sides are given by 4x-7y+10= 0 ,x+y-5=0 and 7x+4y-a5=0 is a) (-1, -2) - Maths - Straight Lines

Question

orthocentre of a triangle whose sides are given by 4x-7y+10= 0
,x+y-5=0 and 7x+4y-a5=0 is ??a) (-1, -2) b) (1, -2) c) (-1, 2)d)
(1, 2)

Priyanka Kedia · Accepted Answer

$L e t A B C b e t h e t r i a n g l e a n d p o int O b e t h e o r t h o c e n t r e o f t h e t r i a n g l e A B C . L e t e q u a t i o n o f s i d e A B i s, 4 x - 7 y + 10 = 0 \Rightarrow 4 x - 7 y = - 10 ..... (1) s i d e B C i s, x + y = 5 ...... (2) s i d e A C i s, 7 x + 4 y = 15 ..... (3) S o l v i n g e q u a t i o n (1) a n d (2), w e g e t x = \frac{25}{11}, y = \frac{30}{11} S o, p o int B w i l l b e (\frac{25}{11}, \frac{30}{11}) A n d, S o l v i n g e q u a t i o n (2) a n d (3), w e g e t x = - \frac{5}{3}, y = \frac{20}{3} S o, p o int C w i l l b e (- \frac{5}{3}, \frac{20}{3}) A n d, S o l v i n g e q u a t i o n (1) a n d (3), w e g e t x = 1, y = 2 S o, p o int A w i l l b e (1, 2) N o w, a l t i t u d e A D i s p e r p e n d i c u l a r t o B C, t h e r e f o r e e q u a t i o n o f A D i s, x - y + k = 0 A D i s p a s \sin g t h r o u g h t h e p o int A (1, 2), s o, 1 - 2 + k = 0 \Rightarrow k = 1 T h u s, e q u a t i o n o f A D = x - y + 1 = 0 ...... (4) A n d, a l t i t u d e B E i s p e r p e n d i c u l a r t o A C, t h e r e f o r e e q u a t i o n o f B E i s, 4 x - 7 y + k = 0 B E i s p a s \sin g t h r o u g h t h e p o int B (\frac{25}{11}, \frac{30}{11}), s o, 4 \times \frac{25}{11} - 7 \times \frac{30}{11} + k = 0 \Rightarrow k = 10 T h u s, e q u a t i o n o f B E = 4 x - 7 y + 10 = 0 ...... (5) S o l v i n g e q u a t i o n (4) a n d (5), w e g e t x = 1, y = 2 T h u s, t h e c o o r d i n a t e o f t h e o r t h o c e n t r e a r e (1, 2) . H e n c e o p t i o n (d) i s c o r r e c t .$

orthocentre of a triangle whose sides are given by 4x-7y+10= 0 ,x+y-5=0 and 7x+4y-a5=0 is ??a) (-1, -2) b) (1, -2) c) (-1, 2)d) (1, 2)