Problem Solving

In a xy-coordinate plane with origin at O)P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by

A. 18* $$\sqrt{2}$$
B. 3* $$\sqrt{2}$$
C. 6* $$\sqrt{2}$$
D. 15*$$\sqrt{2}$$
E. 15*$$\sqrt{2}$$ A

The origin is at 0 which means it is the originating point for x and y axis.
P and Q are the two point on the straight line.
$$3x+4y=-15$$
This gives us a triangle POQ on the straight line $$3x+4y=-15$$ there will be two intersection on x and y axis
$$Let\ x-axis\ inter\sec tion\ =A$$
$$Let\ y-axis\ inter\sec tion\ =B$$
Another perpendicular line from origin O to point C which is in between AB on the line $$3x+4y=-15$$ can divide triangle POQ into two right angle triangles
Line OC is perpendicular to PQ (or AB).This makes it the lenght of triangle AOB and POQ
OA =5 and OB= 15/4 (because points A and B are intersection of the line $$3x+4y=-15$$ with x and y axis respectively.
$$AB=Hypotenuse=\sqrt{OA^2+OB^2}=\sqrt{5^2+\left(\frac{15}{4}\right)^2}=\frac{25}{4}$$
As AOB and OCB as similar then
$$\frac{OC}{OB}=\frac{OA}{OB}$$
$$OC=\frac{OA\cdot OB}{AB}=5\cdot\frac{\frac{15}{4}}{\frac{25}{4}}=\frac{15}{5}=3$$
$$CP=CQ=\sqrt{OP^2}-OC^2=\sqrt{9^2-3^2}=\sqrt{81-8}=\sqrt{72}=\sqrt{36\cdot2}=6\sqrt{2}$$

$$PQ=CP=CQ$$
$$6\sqrt{2}+6\sqrt{2}$$
$$12\sqrt{2}$$
$$Area\ of\ Triangle\ \ OPQ=\frac{1}{2}\cdot b\cdot h$$
$$=\frac{PQ\cdot OC}{2}$$
$$\frac{12\sqrt{2}\cdot3}{2}=18\sqrt{2}$$
$$Answer\ is\ Option\ A$$