Question

If $A$ and $B$ are two points on the line $3x+4y+15=0$ such that $OA=OB=9$ units, then the area of the triangle $OAB$ is

• A. $18$ sq. units
• B. $18\sqrt{2}$ sq. units
• C. $\frac{18}{\sqrt{2}}$ sq. units
• D. None of these

Answer 1

The correct option is B

$18\sqrt{2}$ sq. units

Explanation for the correct option:

Step 1. Find the area of $△OAB$:

Perpendicular distance from $\left(0,0\right)$ to the line $3x+4y+15=0$ is:

$\begin{array}{rcl}d& =& \frac{c}{\sqrt{(a^2+b^2)}}\\ & =& \frac{15}{\sqrt{(3^2+4^2)}}\\ & =& \frac{15}{5}\\ & =& 3\end{array}$

$OA=9$

Base $=2\sqrt{9^2-3^2}$

$$\begin{gathered} =2\sqrt{72} \\ =12\sqrt{2} \end{gathered}$$

Step 2. Area of $\mathbf{△}\mathit{O}\mathit{A}\mathit{B}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}$X BASE X HEIGHT

$$\begin{gathered} =\frac{1}{2}\times 12\sqrt{2}\times 3 \\ =18\sqrt{2} \end{gathered}$$

$\therefore$Area of $△OAB$ is $18\sqrt{2}$ sq. units

Hence, Option ‘B’ is Correct.