Question

If (1, 2) is the orthocentre and (2, -1) is the circumcentre of a triangle ABC, then find the centroid of the triangle ABC​

Answer 1

$\large\underline{\bold{Solution-}}$

Understand the question :-

We know that,

The orthocentre (H), centroid (G) and circumcentre (O) of any triangle are collinear and the centroid (G) divides the distance from the orthocentre (H) to the circumcentre (O)  in the ratio 2:1.

Here,

We have given the orthocentre and circum-centre, and we have to find the coordinates of Centroid. So, we have to use Section Formula to find the coordinates of Centroid.

Formula Used :-

Section Formula :-

Let us assume a line segment joining the points A and B and let C (x, y) be any point which divides the line segment joining AB in the ratio m : n internally, then coordinates of C is given by

$\bf \:( x, y) = \bigg(\dfrac{mx_2 + nx_1}{m + n} , \dfrac{my_2 + ny_1}{m + n} \bigg)$

$\sf \: where \: coordinates \: are \: A(x_1, \: y_1) \: and \: B(x_2, \: y_2)$

Calculations :-

Given that

Coordinates of Orthocentre = H (1, 2)

Coordinates of Circum - centre = O (2, - 1)

Let coordinates of Centroid (G) be (x, y).

Since,

G divides H and O in the ratio 2 : 1,

So,

By using section formula, we have

$\sf \:( x, y) = \bigg(\dfrac{mx_2 + nx_1}{m + n} , \dfrac{my_2 + ny_1}{m + n} \bigg)$

Here,

• x₁ = 1

• x₂ = 2

• y₁ = 2

• y₂ = - 1

• m = 2

• n = 1

Tʜᴜs,

The coordinates of Centroid G (x, y) is,

$\sf \: (x, y) \: = \bigg(\dfrac{2 \times 2 + 1 \times 1}{1 + 2} , \dfrac{ - 1 \times 2 + 2 \times 1}{1 + 2} \bigg)$

$\sf \: (x, y) \: = \bigg(\dfrac{4 + 1}{3} , \dfrac{ - 2 + 2}{3} \bigg)$

$\sf \: (x, y) \: = \bigg(\dfrac{5}{3} , \dfrac{0}{3} \bigg)$

$\sf \: (x, y) \: = \bigg(\dfrac{5}{3} , 0 \bigg)$

Hence,

$\bf \: Coordinates \: of \: Centroid \: is \: \sf \: \bigg(\dfrac{5}{3} , 0 \bigg)$

Additional Information :-

Circumcenter O, :- the point of which is equidistant from all the vertices of the triangle;

Incenter I, :- the point of which is equidistant from the sides of the triangle;

Orthocenter H :- the point at which all the altitudes of the triangle intersect;

Centroid G :-  the point of intersection of the medians of the triangle.

An important relationship between these points is the Euler line, which states that O, G, H is a straight line and OG : GH = 1 : 2.

In fact, the center of the Nine Point Circle is also the midpoint of OH.