Here we are going to see how to find orthocenter of a triangle with given vertices.
What is Orthocenter ?
It can be shown that the altitudes of a triangle are concurrent and the point of concurrence is called the orthocentre of the triangle.The orthocentre is denoted by O.
Let ABC be the triangle AD,BE and CF are three altitudes from A, B and C to BC, CA and AB respectively.
Steps Involved in Finding Orthocenter of a Triangle :
Example 1 :
Find the co ordinates of the orthocentre of a triangle whose vertices are (3, 4) (2, -1) and (4, -6).
Solution :
Let the given points be A (3, 4) B (2, -1) and C (4, -6)
Slope of AC = [(y2 - y1)/(x2 - x1)]
A (3, 4) and C (4, -6)
here x1 = 3, x2 = 4, y1 = 4 and y2 = -6
= [(-6-4)/(4-3)]
= (-10)/1
= -10
Slope of the altitude BE = -1/ slope of AC
= -1/(-10)
= 1/10
Equation of the altitude BE :
(y - y1) = m (x -x1)
Here B (2, -1) and m = 1/10
(y - (-1)) = 1/10 (x - 2)
10 (y + 1) = (x - 2)
10y + 10 = x - 2
x - 10y - 2 - 10 = 0
x - 10y - 12 = 0
x - 10y - 12 = 0
x = 10y + 12 --------(1)
Slope of BC = (y2 - y1)(x2 - x1)]
B (2,-1) and C (4,-6)
here x₁ = 2,x₂ = 4,y₁ = -1 and y₂ = -6
= [(-6 - (-1))/(4 - 2)]
= (-6 + 1)/2
= -5/2
Slope of the altitude AD = -1/ slope of AC
= -1/(-5/2)
= 2/5
Equation of the altitude AD :
(y - y1) = m (x - x1)
Here A(3, 4) m = 2/5
(y - 4) = 2/5 (x - 3)
5(y - 4) = 2(x - 3)
5y - 20 = 2x - 6
2x - 5y - 6 + 20 = 0
2x - 5y + 14 = 0 --------(2)
Substituting (1) into (2), we get
2(10y + 12) - 5y + 14 = 0
20y + 24 - 5y + 14 = 0
15y + 38 = 0
15y = -38
y = -38/15
By applying y = -38/15 in (1), we get
x = 10(-38/15) + 12
x = -76/3 + 12
x = -40/3
Therefore the orthocenter is (-40/3, -38/15).
After having gone through the stuff given above, we hope that the students would have understood how to find the orthocenter of a triangle with given vertices.
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