Question

Centroid of scalen triange using dormula

Answer 1

Let ∆ ABC be the given triangle.

Perpendiculars BE on AC and CF on AB intersect at O.

Construction: Join AO and produce it to meet BC at D.

Join EF.

In order to prove that the perpendicular from the vertices on opposite sides are concurrent, it is sufficient to prove that AD ⊥ BC.



BE ⊥ AC and CF ⊥ AB

⇒ ∠BEC = 90°  and ∠BFC = 90°

⇒ ∠BEC = ∠BFC = 90°

⇒ BC makes equal angles at E and F.

⇒ Points B, C, E, F are concyclic.

⇒ BCEF is a cyclic quadrilateral.

⇒ ∠ECB + ∠BFE = 180°

⇒ ∠ECB  + (∠BFC + ∠CFE)= 180°

⇒ ∠ECB + 90° + ∠CFE = 180°

⇒ ∠ACD +∠OFE = 90°  ............... (1)

In quadrilateral EOFA,

⇒ ∠OFA + OEA + = 90° + 90° = 180°

⇒ EOFA is a cyclic quadrilateral.

⇒ ∠ OFE = ∠OAE  ........... (2)  [ Angles in the same segment are equal ]

from (1) and (2)

 ∠ACD + ∠OAE = 90°

⇒ ∠ACD + ∠DAE = 90°

⇒ ∠ACD + ∠DAC = 90°

⇒ 180° – ∠ADC = 90° [ Since, in ∆ADC, ∠ACD + ∠ADC + ∠DAC = 180° ]

⇒ ∠ADC = 90°

⇒ AD ⊥ BC.

Hence, altitude AD, BE and CF are concurrent

Answer 2

hi friend!!

If (x1,y1) ,(x2,y2) and (x3,y3) are the vertices of the scalene triangle.

Then centroid of the triangle will be given by the formula

(x1+x2+x3/3,y1+y2+y3/3)


I hope this will help u;)