Question

Find the area of triangle $OPB$ in the above figure given that point $O$ has coordinates $(0,0)$, point $P$ lies on the graph of $y=6-x^2$, and the point $B$ has coordinates $(2\sqrt{3},0)$. Also $l\left(OP\right)=l\left(BP\right)$.

• A. $1.7$
• B. $3.0$
• C. $3.5$
• D. $4.7$
• E. $5.2$

Answer 1

The correct option is E $5.2$

The $x$-coordinate of $P$ will be midpoint of $O$ and $B$, which implies $x$-coordinate of $P$ is $\sqrt{3}$

Therefore $y$-coordinate is $6-3=3$

The area of triangle with vertices $(0,0)$, $(\sqrt{3},3),(2\sqrt{3},0)$ will be $\frac{1}{2}\times |0+\sqrt{3}\left(0\right)+2\sqrt{3}(0-3)|$

$=\frac{1}{2}\times 6\sqrt{3}$

$=3\sqrt{3}$

$=5.2$