Question

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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Answer 1

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 2

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 =  $[\frac{x_{1}+ x_{2} +x_{3} }{3}, \frac{y_{1}+ y_{2} +y_{3} }{3} ]$

⇒ G $[\frac{x_{1}+ x_{2}+ x_{3} }{3} , \frac{y_{1} +y_{2} + y_{3} }{3} ]$  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

       = $[\frac{mx_{2} + nx_{1} }{ m+n} , \frac{my_{2}+ny_{1} }{m+n} ]$  

⇒ the point which divides O (P,Q) and C (0,0)  in 2 : 1

    = $[\frac{ 2(0) + 1(P)}{2+1} , \frac{2(0)+ 1(Q)}{2+1} ]$  which will be equal to G $[\frac{x_{1}+ x_{2}+ x_{3} }{3}, \frac{y_{1} +y_{2} +y_{3} }{3} ]$  

⇒    $[\frac{0+P}{3} , \frac{0+Q}{3} ]$ = $[\frac{x_{1} +x_{2} +x_{3} }{3} , \frac{y_{1} +y_{2}+ y_{3} }{3} ]$

⇒  $\frac{P}{3} = \frac{x_{1}+ x_{2} +x_{3} }{3}$                       ⇒  $\frac{Q}{3} = \frac{y_{1}+ y_{2} +y_{3} }{3}$    

⇒  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)