Ans: (a) (i) $x=11$ (ii) $(11,23)$ (b) (i) $6$ (ii) $69$
-
- $x=11$.
- $(11,23)$.
-
- Sub. $y=5$ into $y=-2(x-11)^2 + 23$, we have
$\begin{array}{rcl}
-2(x-11)^2 + 23 & = & 5 \\
-2x^2 + 44x -224 & = & 0 \\
x^2 – 22x + 112 & = & 0 \\
(x-8)(x-14) & = & 0
\end{array}$Therefore, $x=8$ or $x=14$. Hence, the distance between $P$ and $Q$
$\begin{array}{cl}
= & 14 – 8 \\
= & 6
\end{array}$ - Note that $PQ$ is a horizontal line. Then the distance between $R$ and $PQ$
$\begin{array}{cl}
= & 23 – 5 \\
= & 18
\end{array}$Therefore, the area of $\Delta RPQ$
$\begin{array}{cl}
= & \dfrac{1}{2} \times 18 \times 6 \\
= & 54
\end{array}$And the distance between $PQ$ and the $x$-axis
$\begin{array}{cl}
= & 5 – 0 \\
= & 5
\end{array}$Therefore, the area of $\Delta PQS$
$\begin{array}{cl}
= & \dfrac{1}{2} \times 5 \times 6 \\
= & 15
\end{array}$Hence, the area of quadrilateral $PQRS$
$\begin{array}{cl}
= & 54 + 15 \\
= & 69
\end{array}$
- Sub. $y=5$ into $y=-2(x-11)^2 + 23$, we have
