91Ó°ÊÓ

Given figure is:

We have to show that $\mathrm{Δ}OABâ‰\dots \mathrm{Δ}ORS$.

In the given figure:

In $\mathrm{Δ}OAB$$and\mathrm{Δ}ORS$

$\begin{array}{c}\stackrel{¯}{OA}=\stackrel{¯}{OR}\\ \stackrel{¯}{AB}=\stackrel{¯}{RS}\\ ∠A=∠R\\ \mathrm{Δ}OABâ‰\dots \mathrm{Δ}ORS\end{array}$

$So,\mathrm{Δ}OABâ‰\dots \mathrm{Δ}ORS$

Given figure is:

We have to find why$\stackrel{¯}{OB}âŠ\yen \stackrel{¯}{OS}$ .

In the given figure:

Let us assume that$∠BOA={\text{n}}^{\text{o}}$

Then,$∠ROB={90}^{\text{o}}−{\text{n}}^{\text{o}}$

From part a,$\mathrm{Δ}OABâ‰\dots \mathrm{Δ}ORS$.

So,$∠BOE=∠FNE\left(=n^{\text{o}}\right)$

Then we calculate the angle$∠SOB$,

$\begin{array}{l}∠SOB=∠ROS+∠ROB\\ ∠SOB=n^{\text{o}}+{90}^{\text{o}}−n^{\text{o}}\\ ∠SOB={90}^{\text{o}}\end{array}$

Therefore,$\stackrel{¯}{OB}âŠ\yen \stackrel{¯}{OS}$

Given figure is:

We have to find the product of the slopes of $\stackrel{¯}{OB}$and $\stackrel{¯}{OS}$.

In the given figure:

Given coordinates of$\stackrel{¯}{OB}$are (0, 0) and (5, 3).

Then we calculate the slope of the line

$m=\frac{y_2−y_1}{x_2−x_1}$

Substituting the values for the slope of $\stackrel{¯}{OB}$.

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

The coordinates of are (0, 0) and (3, 5).

Slope of the$\stackrel{¯}{OS}$

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

Then we calculate the product of the slopes.

$\begin{array}{l}P=\frac{3}{5}×\frac{5}{3}\\ P=1\end{array}$

So, the product of the slopes of and is .