$A(-3,-3),B(2,-2),C(2,3)$ and $D(-4,4)$ are the vertices of quadrilateral $ABCD$. Then the point of intersection of its diagonals is

The correct option is C. $(\frac{-3}{11},\frac{3}{11})$

We know that the slope(m) of the line formed by joining the points $(x_1,y_1)$ and $(x_2,y_2)$ is given by,

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

Now, the slope of the diagonal AC

= $\frac{3-(-3)}{2-(-3)}$

=$\frac{6}{5}$

And the slope of the diagonal BD is

=$\frac{4-(-2)}{-4-2}$

=$-1$

Let $P(x,y)$ be the point of intersection of these diagonals.

Then $P(x,y)$ lies on both AC and BD.

Therefore $\frac{y-3}{x-2}=\frac{6}{5}$ and $\frac{y-(-2)}{x-2}=-1$

$⟹5y-15=6x-12$ and $y+2=-x+2$

$⟹6x-5y+3=0$ and $x+y=0$

Solving the above two equations, we get, $x=\frac{-3}{11}$ and $y=\frac{3}{11}$

Hence the point of intersection P of the diagonals AC and BD is $(\frac{-3}{11},\frac{3}{11})$.