Question

The are of quadrilateral formed by \[5x \pm2y \pm5=0\]

Answer 1

I got 5 square units

Answer 2

u r correct

Answer 3

Can u shw ???

Answer 4

These are the coordinate points of the given quadrilateral

Answer 5

Using application of determinant for area we can find out the answer

Answer 6

Lol...we will be having four diff eq

Answer 7

Ok here is it
\[\frac{ 1 }{ 2 }|\left[\begin{matrix}0 & \frac{ 5 }{ 2 } \\ 1 & 0\end{matrix}\right]+\left[\begin{matrix}1 & 0 \\ 0 & \frac{ -5 }{ 2 }\end{matrix}\right]+\left[\begin{matrix}0 & \frac{ -5 }{ 2 }\ \\ -1 & 0\end{matrix}\right]+\left[\begin{matrix}-1 & 0 \\ 0 & \frac{ 5 }{ 2 }\end{matrix}\right]|\]

Answer 8

thxx

Answer 9

@Zekarias sorry But I didnt get what u did?

Answer 10

That was the application of determinant