Question

The centroid of triangle is (1, 4) and coordinates of its two vertices are (4, - 3) and -(9, 7) then, two-third of the area of triangle is _________. (in sq. units

Answer 1

Hi friend!!!

Given, The centroid of triangle is (1, 4) and coordinates of its two vertices are (4, - 3) and (9, 7)

let the third vertice be (x,y)

then ({4+9+x}/3,{-3+7+y}/3)=(1,4)

By comparing x- coordinate and y coordinate we get 13+x=3→x=-10

4+y=12→y=8

so, third vertex is (-10,8)

we know that the area of a triangle whose vertices are (a,b),(c,d),(e,f) is

1/2.{a(d-f)+c(f-b)+e(b-d)}

here a=4 b=-3 c=9 d=7 e= -10 f=8

by substituting..

Area=1/2{4(7-8)+9(8+3)-10(-3-7)}

=1/2{4(-1)+9(11)-10(-10)}
=1/2{-4+99+100}

=195/2

Now, 2/3 of the area = 2/3 × 195/2
=65sq.units

I hope this will help u ;)

Happy learning