91Ó°ÊÓ

Given figure is:

We have to find the slopes of$\stackrel{¯}{OD}$ and$\stackrel{¯}{NF}$ .

In the figure (1):

Given coordinates of$OD$are (0, 0) and (3, 3).

We have to calculate the slope of the line:

$m=\frac{y_2−y_1}{x_2−x_1}$

Substituting the values in the formula of slope for$\stackrel{¯}{OD}$.

$\begin{array}{l}m=\frac{3−0}{3−0}\\ m=\frac{3}{3}\\ m=1\end{array}$

Given that the coordinates of $\stackrel{¯}{NF}$are (5, 0) and (8, 3).

We will calculate the slope of$\stackrel{¯}{NF}$.

$\begin{array}{l}m=\frac{3−0}{8−5}\\ m=\frac{3}{3}\\ m=1\end{array}$

So, the slope of$\stackrel{¯}{OD}$ and$\stackrel{¯}{NF}$ is $1$.

Given figure is:

We have to show that $\mathrm{Δ}OCDâ‰\dots \mathrm{Δ}NEF$.

In $\mathrm{Δ}OCD$and $\mathrm{Δ}NEF$in the given figure:

$\begin{array}{l}\stackrel{¯}{OC}=\stackrel{¯}{NE}\\ \stackrel{¯}{CD}=\stackrel{¯}{EF}\\ ∠C=∠E\left({\text{90}}^{\text{o}}\right)\end{array}$

So, $\mathrm{Δ}OCDâ‰\dots \mathrm{Δ}NEF$

Hence, $\mathrm{Δ}OCDâ‰\dots \mathrm{Δ}NEF$by SAS rule.

Given figure is:

We have to show that $∠DOCâ‰\dots ∠FNE$.

From the given figure:

The slope of$\stackrel{¯}{OD}$ is equal to the slope of$\stackrel{¯}{NF}$

So,$\tan ∠DOC=\tan ∠FNE$

Hence,$∠DOCâ‰\dots ∠FNE.$

Given figure is:

We have to find why $\stackrel{¯}{OD}âˆ\yen \stackrel{¯}{NF}$.

In the given figure:

We see that$\stackrel{¯}{OD}$and $\stackrel{¯}{NF}$have the same angle with the , and it is also evident from the above solution.

$\tan ∠DOC=\tan ∠FNE$

It means $\stackrel{¯}{OD}âˆ\yen \stackrel{¯}{NF}$

So,$\stackrel{¯}{OD}âˆ\yen \stackrel{¯}{NF}$

Given figure is:

We have to find the slope of parallel lines.

In the given figure,

The parallel lines have the same angle with the .

So, the slopes of the parallel lines are the same.