Question

A(1,4), B(3,2)and C(7,5) are the vertices of a triangle ABC.Find:

(1) the coordinates of the centroid G of triangle ABC.

(2) the equation of a line through G and parallel to AB.

Answer 1

Answer :

Points A ( 1 , 4 ) , B ( 3 , 2 ) And C ( 7 , 5 ) are the vertices of triangle ABC

1 ) We know coordinates of centroid are $\left(\frac{x_1+x_2+\hspace{0.17em}{x}_3}{2},\frac{y_1+y_2+\hspace{0.17em}{y}_3}{2}\right)$
Here x₁ = 1 , x₂ = 3 And x₃ = 7
y₁ = 4 , y₂ = 2 And y₃ = 5
So,

Coordinates of centroid G = $\left(\frac{1+3+\hspace{0.17em}7}{2},\frac{4+2+\hspace{0.17em}5}{2}\right)$

G = $\left(\frac{11}{2},\frac{11}{2}\right)$ ( Ans )

2 ) Now we find the sloe of equation AB , As

Slope = $\frac{y_1-y_2}{x_1-x_2}$

here​ x₁ = 1 , x₂ = 3 And ​y₁ = 4 , y₂ = 2
So slope of AB = $\frac{4-2}{1-3}$ = -1
And we know if two lines are parallel to each other than they have same slope ,
And
we know equation of line that passing through a point = ( y - y₁ ) = m ( x - x₁ )
so here x₁ = $\frac{11}{2}$ And y₁ = $\frac{11}{2}$ as our line passing through centoid G ( $\frac{11}{2}$ , $\frac{11}{2}$ )

So equation of line parallel to AB and passing through G , is

( y - $\frac{11}{2}$ ) = ( - 1 ) ( x - $\frac{11}{2}$ )

y - $\frac{11}{2}$ = - x + $\frac{11}{2}$

x + y = $\frac{11}{2}$ + $\frac{11}{2}$

x + y = 11 ( Ans )