Question

A(-2,-1), B(0,-3), C(1,0) and D(-1,1) are the vertices of quadrilateral ABCD. Then the point of intersection of its diagonals is

• A. $(\frac{8}{13},\frac{-7}{13})$
• B. $(\frac{-8}{13},\frac{-7}{13})$
• C. $(\frac{-3}{11},\frac{7}{13})$
• D. $(\frac{3}{13},\frac{7}{11})$

Answer 1

The correct option is B

$(\frac{-8}{13},\frac{-7}{13})$

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

$=\frac{0-(-1)}{1-(-2)}$

$=\frac{1}{3}$

And the slope of the diagonal BD is

$=\frac{1-(-3)}{-1-0}$

$=-4$

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

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

Therefore, $\frac{y-0}{x-1}=\frac{1}{3}and\frac{y-1}{x-(-1)}=-4$

$⟹3y=x-1andy-1=-4x-4$

$⟹x-3y=1and4x+y=-3$

Solving the above two equations, we get, $x=\frac{-8}{13}$ and $y=\frac{-7}{13}$

Hence the point of intersection P of the diagonals AC and BD is $(\frac{-8}{13},\frac{-7}{13})$