The centroid and two vertices of a Triangles are (4,-8),(-9,7 1,4) then the area of the triangle is

1) 95/6

2) 285/2

3) 190/3

4) 285

Solution:

We know the centroid of a triangle is {(x1+x2+x3)/3, (y1+y2+y3)/3}

Here centroid is (0, 0) and two vertices are (-8, 7) and (9, 4).

So (x1 – 8 + 9)/3 = 0 and (y1 + 7 + 4) = 0

=> x1 + 1 = 0 and y1 + 11 = 0

=> x1 = -1, y1 = -11

So the third vertex is (-1, -11)

Area of triangle ∆ =

\(\begin{array}{l}\frac{1}{2}\begin{vmatrix} x_{1} & y_{1} & 1\\ x_{2}&y_{2} & 1\\ x_{3}& y_{3} & 1 \end{vmatrix}\end{array} \)

=

\(\begin{array}{l}\frac{1}{2}\begin{vmatrix} -1 & -11 & 1\\ -8&7 & 1\\ 9& 4 & 1 \end{vmatrix}\end{array} \)

= |(½)[ -1(7 – 4) + 11(-8 -9) + 1(-32 – 63)]|

= |(½)[ -3 – 187 – 95]|

= |-285/2|

= 285/2

Hence option (2) is the answer.