A(2, 5), B(6, – 1), C(- 4, – 3) are the vertices of ∆ABC.
Let AD, BE and CF be the altitudes through the vertices A, B and C respectively of ∆ABC.
∴ Slope of AD = -5 …[∵AD ⊥ BC]
Since altitude AD passes through (2, 5) and has slope – 5,
equation of the altitude AD is y – 5 = -5 (x – 2)
∴ y – 5 = – 5x + 10
∴ 5x +y -15 = 0
Now, slope of AC = -3-5/ -4-2 = -8/-6 = 4/3
Slope of BE = -3/4 …[∵ BE ⊥ AC]
Since altitude BE passes through (6,-1) and has slope -3/4
equation of the altitude BE is y- (-1) = -3/4 (x – 6)
∴ 4 (y + 1) = – 3 (x – 6)
∴ 4y + 4 =-3x+ 18
∴ 3x + 4y – 14 = 0
Also, slope of AB = -1-5/ 6-2 = -6/4 = -3/2
∴ Slope of CF = ….[∵ CF ⊥ AB]
Since altitude CF passes through (- 4, – 3) and has slope , 2/3
equation of the altitude CF is y - (-3) = 2/3 [x-(-4)]
∴ 3 (y + 3) = 2 (x + 4)
∴ 3y + 9 = 2x + 8
∴ 2x – 3y – 1 = 0
