If A and B are two points on the line 3x + 4y | vedclass

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(B) Let $O$ be the origin $(0, 0)$. The perpendicular distance $OD$ from the origin to the line $3x + 4y + 15 = 0$ is given by $OD = \frac{|3(0) + 4(0

Vedclass · Accepted Answer

$18\sqrt{2} 	ext{ sq. units}$

Vedclass · Answer

(B) Let $O$ be the origin $(0, 0)$. The perpendicular distance $OD$ from the origin to the line $3x + 4y + 15 = 0$ is given by $OD = \frac{|3(0) + 4(0) + 15|}{\sqrt{3^2 + 4^2}} = \frac{15}{5} = 3 	ext{ units}$.In the right-angled triangle $ODA$,we have $OA = 9$ and $OD = 3$.Using the Pythagorean theorem,$AD = \sqrt{OA^2 - OD^2} = \sqrt{9^2 - 3^2} = \sqrt{81 - 9} = \sqrt{72} = 6\sqrt{2} 	ext{ units}$.Since $OD$ is perpendicular to $AB$,$D$ is the midpoint of $AB$,so $AB = 2AD = 2(6\sqrt{2}) = 12\sqrt{2} 	ext{ units}$.The area of triangle $OAB$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AB 	imes OD = \frac{1}{2} 	imes 12\sqrt{2} 	imes 3 = 18\sqrt{2} 	ext{ sq. units}$.

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

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