If the centroid of a triangle is (1, 4)

Solution:

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 $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)$ is

$\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$

So,

$(1,4)=\left(\frac{x+4-9}{3}, \frac{y-3+7}{3}\right)$

Compare individual terms on both the sides-

$\frac{x-5}{3}=1$

So,

$x=8$

Similarly,

$\frac{y+4}{3}=4$

So,

$y=8$

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

In general if $\mathrm{A}\left(x_{1}, y_{1}\right) ; \mathrm{B}\left(x_{2}, y_{2}\right) ; \mathrm{C}\left(x_{3}, y_{3}\right)$ are non-collinear points then are of the triangle formed is given by-,

$\operatorname{ar}(\Delta \mathrm{ABC})=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$

So,

$\operatorname{ar}(\triangle \mathrm{ABC})=\frac{1}{2}|4(7-8)-9(8+3)+8(-3-7)|$

$=\frac{1}{2}|-4-99-80|$

$=\frac{183}{2}$

So the answer is (b)