91Ó°ÊÓ

Let x denotes the number of sides and the corresponding number of diagonals.

For $x=3,y=0$and for $x=4,y=2$

So, take the two points as $\left(3,0\right)$and $\left(4,2\right)$

Let $\left(x_1,y_1\right)=\left(3,0\right)$and $\left(x_2,y_2\right)=\left(4,2\right)$.

Equation of a line passing through points $\left(x_1,y_1\right)$and$\left(x_2,y_2\right)$ is given by

$y-y_2=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$ …… (1)

Substitute $\left(4,2\right)$for$\left(x_2,y_2\right)$ and$\left(3,0\right)$ for$\left(x_1,y_1\right)$ in equation (1) as follows:

$$\begin{gathered} y-2=\frac{2-0}{4-3}\left(x-3\right) \\ y-2=2\left(x-3\right) \end{gathered}$$ …… (2)

Now find the number of diagonals y corresponding to the given number of sides x as:

Calculate for $x=5$ the value for y as:

$$\begin{gathered} y-2=2\left(x-3\right) \\ y=\frac{2\left(5-3\right)}{2} \\ y=2 \end{gathered}$$

Calculate for$x=6$ the value for y as:

$$\begin{gathered} y-2=2\left(x-3\right) \\ y=\frac{2\left(6-3\right)}{2} \\ y=3 \end{gathered}$$

Calculate for$x=7$ the value for y as:

$$\begin{gathered} y-2=2\left(x-3\right) \\ y=\frac{2\left(7-3\right)}{2} \\ y=4 \end{gathered}$$

Calculate for$x=8$the value for y as:

$$\begin{gathered} y-2=2\left(x-3\right) \\ y=\frac{2\left(8-3\right)}{2} \\ y=5 \end{gathered}$$

Now,calculate for$x=n$the value for y as:

$$\begin{gathered} y-2=2\left(x-3\right) \\ y=\frac{2\left(n-3\right)}{2} \\ y=\left(n-3\right) \end{gathered}$$

Therefore, the complete table is