In the Δ ABC shown below,
AD ⊥ BC, BE ⊥ AC, CF ⊥ AB
The line segments AD, BE and CF are called altitudes of the triangle ABC.
Since AD ⊥ BC, product of slopes of AD and BC is equal to -1.
slope of AD x slope of BC = -1
slope of AD = -1/slope of BC
Similarly,
slope of BE = -1/slope of AC
slope of CF = -1/slope of AB
Example :
Let A(1 , 2), B(-4 , 5) and C(0 , 1) be the vertices of the Δ ABC. Find the slopes of the altitudes AD, BE and CF.
Solution :
Slope of AD :
Use slope formula and find the slope of BC.
m = (y_{2 }- y_{1})/(x_{2 }- x_{1})
Substitute (x_{1}, y_{1}) = B(-4, 5) and (x_{2}, y_{2}) = C(0, 1).
slope of BC = (1 - 5)/(0 + 4)
= -4/4
= -1
slope of AD = -1/slope of BC
= -1/(-1)
= 1
Slope of BE :
m = (y_{2 }- y_{1})/(x_{2 }- x_{1})
Substitute (x_{1}, y_{1}) = A(1, 2) and (x_{2}, y_{2}) = C(0, 1).
slope of AC = (1 - 2)/(0 - 1)
= -1/(-1)
= 1
slope of BE = -1/slope of AC
= -1/1
= -1
Slope of CF :
m = (y_{2 }- y_{1})/(x_{2 }- x_{1})
Substitute (x_{1}, y_{1}) = A(1, 2) and (x_{2}, y_{2}) = B(-4, 5).
slope of AB = (5 - 2)/(-4 - 1)
= 3/(-5)
= -3/5
slope of CF = -1/slope of AB
= -1/(-3/5)
= 5/3
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