a. Show that$\Delta OAB\cong \Delta ORS$.

b. Why is$\overline{OB}\perp \overline{OS}$?

c. Find the product of the slopes of$\overline{OB}$and$\overline{OS}$.

Given figure is:

We have to show that $\Delta OAB\cong \Delta ORS$.

In the given figure:

In $\Delta OAB$$and\Delta ORS$

$\begin{array}{c}\overline{OA}=\overline{OR}\\ \overline{AB}=\overline{RS}\\ \angle A=\angle R\\ \Delta OAB\cong \Delta ORS\end{array}$

$So,\Delta OAB\cong \Delta ORS$

Given figure is:

We have to find why$\overline{OB}\perp \overline{OS}$ .

In the given figure:

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

Then,$\angle ROB={90}^{\text{o}}-{\text{n}}^{\text{o}}$

From part a,$\Delta OAB\cong \Delta ORS$.

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

Then we calculate the angle$\angle SOB$,

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

Therefore,$\overline{OB}\perp \overline{OS}$

Given figure is:

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

In the given figure:

Given coordinates of$\overline{OB}$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 $\overline{OB}$.

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

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

Slope of the$\overline{OS}$

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

Then we calculate the product of the slopes.

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

So, the product of the slopes of and is .