Question

If (2,3) is the vertex and D (1,-2) is the mid of of BC of triangle ABC.Then find the centroid the triangle ABC.​

Answer 1

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

We know,

↝ Median is a line segment drawn from the vertex to bisects the opposite sides and medians divides each other in the ratio 2 : 1.

Since,

It is given that,

↝ In triangle ABC,

• Coordinates of vertex A = (2, 3)

and

• Coordinates of D = (1, - 2),

• where D is the midpoint of BC.

So,

↝ It implies AD is median of triangle ABC.

↝ Let assume that Centroid of a triangle be 'G' having coordinates (x, y).

Now,

↝ G divides AD in the ratio 2 : 1.

We know,

Section Formula is used to find the coordinates of the point (x, y) which divides the line segment joining the points A and B in the ratio m : n internally, and coordinates of C is given by

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

So, According to given data, we have

$\rm :\longmapsto\:x_1 =2 , \: y_1 = 3, \: x_2=1 , \: y_2= - 2, \: m = 2, \: n = 1$

So, on substituting the values we get

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

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

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

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

Hence,

Coordinates of Centroid G (x, y) is

$\rm :\longmapsto\:\bf{ \:( x, y) = \bigg(\dfrac{4}{3} , \dfrac{ - 1}{3} \bigg)}$

Additional Information :-

Distance Formula :-

$\red{ \boxed{ \sf{ \:Distance = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }}}}$

Midpoint Formula :-

$\red{ \boxed{ \sf{ \:\:( x, y) = \bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} \bigg) }}}$

Area of triangle :-

$\red{ \boxed{ \sf{Area =\dfrac{1}{2} [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]}}}$

Condition for 3 points to be collinear

$\red{ \boxed{ \sf{ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) = 0}}}$