Question

If P, Q are two points on the line $3x+4y+15=0$ such that OP+PQ=9, then area of $△OPQ$

Answer 1

We have,

Equation of line is $$3x+4y+15=0......\left(1\right)$$

Let the points A and B are intersection of the line $$3x+4y+15=0$$ with the X and Y axis respectively.

Then,

If point A the X-axis so, y=0

Then,

$3x+4y=-15$

$3\times 0+4y=-15$

$y=\frac{-15}{4}$

If point B the Y- axis so, x=0

Then,

$3x+4y=-15$

$3x+4\times 0=-15$

$x=-5$

Then,

Hypotenuse AB

$=\sqrt{O{A}^2+O{B}^2}$

$=\sqrt{5^2+{\left(\frac{15}{4}\right)}^2}$

$=\sqrt{25+\frac{225}{16}}$

$=\sqrt{\frac{400+225}{16}}$

$=\sqrt{\frac{625}{16}}$

$=\frac{25}{4}$

As $\Delta AOB$ and $\Delta OCB$ are similar then,

$\frac{OC}{OB}=\frac{OA}{AB}$

$\Rightarrow \frac{OC}{\frac{15}{4}}=\frac{5}{\frac{25}{4}}$

$\Rightarrow OC=3$

Now,

$CP=CQ=\sqrt{O{P}^2-O{C}^2}$

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

$=6\sqrt{2}$

$PQ=CP+CQ$

$Now,Ar\left(\Delta OPQ\right)=\frac{1}{2}\times base\times height$

$=\frac{1}{2}\times PQ\times OC$

$=\frac{1}{2}\times 12\sqrt{2}\times 3$

$=18\sqrt{2}$

Hence, this is the answer.