Let O be the orthocentre of ∆ABC.
Let AD and BE be the altitudes on the sides BC and AC respectively.
Slope of side BC = 2-6/-1-7 = -4/-8 = 1/2
∴ Slope of AD = – 2 [∵ AD ⊥ BC]
∴ Equation of line AD is y – (-2) = (- 2) (x – 3)
∴ y + 2 = -2x + 6
∴ 2x + y -4 = 0 …(i)
Slope of side AC = -2-2/ 3-(-1) = -4/-4 = -1
∴ Slope of BE = 1 …[ ∵ BE ⊥ AC]
∴ Equation of line BE is y – 6 = 1(x – 7)
∴ y – 6 = x – 1
∴ x = y + 1 …(ii)
Substituting x = y + 1 in (i), we get 2(y + 1) + y – 4 = 0
∴ 2y + 2 + y – 4 = 0
∴ 3y – 2 = 0
∴ y = 2/3 in (ii) we get x = 2/3 + 1 = 5/3
∴ Co-ordinates of orthocentre, O = (5/3, 2/3)
