Question

If the centroid of a triangle is (1, 4) and two of its vertices are (4, −3) and (−9, 7), then the area of the triangle is

Options

• 183 sq. units

• \[\frac{183}{2}\]  sq. units

   

• 366 sq. units

• \[\frac{183}{4}\]  sq. units

   

   

Answer 1

We have to find the co-ordinates of the third vertex of the given triangle. Let the co-ordinates of the third vertex be ( x , y) .

The co-ordinates of other two vertices are (4,−3) and (−9, 7)

The co-ordinate of the centroid is (1, 4)

We know that the co-ordinates of the centroid of a triangle whose vertices are  `(x_1 ,y_1 ) , (x_2,y_2),(x_3,y_3)`  is

`((x_1+x_2 +x_3)/3 , (y_1 + y_2+y_3)/3)`

So,

`(1 , 4)  = ((x+4-9)/3 , (y-3+7)/3)`

Compare individual terms on both the sides- `(x - 5)/3 = 1`

So,

x = 8

Similarly,

`(y+ 4 )/3 = 4`

So,

y = 8

So the co-ordinate of third vertex is (8, 8)

In general if `A (x_1 , y_1) ;B(x_2 , y_2 ) ;C(x_3 , y_3)`  are non-collinear points then are of the triangle formed is given by-,

`ar (Δ ABC ) = 1/2 |x_1(y_2 - y_3 ) +x_2 (y_3 - y_1) + x_3 (y_1 - y_2)|`

So,

`ar (ΔABC ) = 1/2 |4(7-8)-9(8+3)+8(-3-7)|`

                   `= 1/2 | -4-99-80|`

                   `= 183/2`