if-the-equations-of-the-sides-of-the-triangle-are-7x-y-10-0-x-2y-5-0-and-x-y-2-0-find-the-orhocentre-of-the-triangle-

Question Number 808 by sai dinesh last updated on 16/Mar/15

i f t h e e q u a t i o n s o f t h e s i d e s o f t h e t r i a n g l e a r e 7 x + y - 10 = 0, x - 2 y + 5 = 0 a n d x + y + 2 = 0, f i n d t h e o r h o c e n t r e o f t h e t r i a n g l e

Commented by prakash jain last updated on 16/Mar/15

i f t h e e q u a t i o n s o f t h e s i d e s o f t h e t r i a n g l e a r e

7 x + y - 10 = 0, x - 2 y + 5 = 0 a n d x + y + 2 = 0,

f i n d 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 .

Answered by prakash jain last updated on 16/Mar/15

Lines

y = - 7 x + 10 \dots (i)

y = \frac{1}{2} x + \frac{5}{2} \dots (ii)

y = - x - 2 \dots (iii)

Vertices

(i) and (ii)

x_{1} = 1, y_{1} = 3

(ii) and (iii)

x_{2} = - 3, y_{2} = 1

(iii) and (i)

x_{3} = 2, y_{3} = - 4

Equation of line perpendicular to (ii)

Slope of a perpendicular line - ve reciprocal

y = - 2 x + c

Altitude on line (ii) passes thru (x_{3}, y_{3}) = (2, - 4)

Equation of altitude on (ii)

y = - 2 x \dots . . (altitude 1)

Equation of line perpendicular to (iii)

y = x + c

Altitude on line (iii) passes thru (x_{1}, y_{1}) = (1, 3)

Equation of altitude on (iii)

y = x + 2 \dots . . (altitude 2)

Intersection of altitudes gives orthocenter

x = - \frac{2}{3}, y = \frac{4}{3}