If the circumcentre of a triangle with vertices (x1,y1),(x2,y2),(x3,y3) is (0,0) then prove that the orthocentre is (x1+x2+x3,y1+y2+y3)
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Answers
Let in ▲ABC,
- A (x1, y1)
- B (x2, y2)
- C (x3, y3)
Then,
The abscissa of the orthocentre is given by,
X=[( x1 tan A + x2 tan B + x3 tan C)/(tan A + tan B + tan C)] .
The ordinate of the orthocentre is given by,
Y=[(y1 tan A + y2 tan B + y3 tan C)/(tan A + tan B + tan C)] .
So the orthocentre is (X, Y)
NOTE:
This is a general formula only to be used when values of tan A, tan B and tan C are manually computable or provided along with question.
tanA can be found by calculating the angle between the lines AB and AC. Similarly tan B and tan C can be calculated.
Answer:
As shown below
Step-by-step explanation:
Given vertices of a triangle (x₁, y₁) (x₂, y₂) (x₃, y₃)
circumcenter is (0, 0)
Here we need to prove that orthocenter of the triangle is
(x₁+x₂+x₃, y₁+y₂+y₃)
⇒ In a triangle orthocenter, centroid and cicumcenter are lie on a same line and centroid will divide orthocenter and circumcenter in 2 : 1 ratio
⇒ by using above relation we will calculate the orthocenter of the triangle
⇒ let O be the Orthocentre, G be the Centroid and C be the circumcenter and these 3 points are lies on same line makes a line segment OGC
⇒ let O (P,Q) be the orthocentre of the triangle
⇒ centrode of the triangle with vertices ( x₁, y₁) (x₂,y₂) (x₃, y₃)
G =
⇒ G will divide O (P, Q) and C (0,0) in 2 : 1 ratio
⇒ we know the formula for a point which divides 2 points in m : n ratio
=
⇒ the point which divides O (P,Q) and C (0,0) in 2 : 1
= which will be equal to G
⇒ =
⇒ ⇒
⇒ P = x₁+ x₂+ x₃ ⇒ Q = y₁+y₂+y₃
⇒ G (P, Q) = (x₁+x₂+x₃ , y₁+y₂+y₃)
⇒Hence it is proven that Orthocenter of the triangle is (x₁+x₂+x₃, y₁+y₂+y₃)