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

Perpendicular distance from origin on the line $3x+4y+15=0$ is:

$d=\frac{15}{\sqrt{3^2+4^2}}=\frac{15}{5}=3OA=9unitsBase=2\times \sqrt{9^2-3^2}=2\sqrt{72}=12\sqrt{2}Area=\frac{1}{2}\times base\times height=\frac{1}{2}\times 12\sqrt{2}\times 3=18\sqrt{2}units$