Question
Quadrilateral OABC is shown on the following set of axes.
$OABC\text{ is symmetrical about }[OB].$
$A\text{ has coordinates }(6,0)\text{ and }C\text{ has coordinates }(3,3\sqrt{3}).$
$(a)\quad(i)\text{ Write down the coordinates of the midpoint of }[AC].$
$\quad\quad(ii)\text{ Hence or otherwise, find the equation of the line passing through the points }O\text{ and }B.$
$(b)\text{ Given that }[OA]\text{ is perpendicular to }[AB],\text{ find the area of the quadrilateral }OABC.$
▶️Answer/Explanation
Detailed solution
Part (a)(i): Finding the coordinates of the midpoint of \([AC]\)
We’re given the coordinates of points \(A\) and \(C\):
– \(A\) is at \((6, 0)\),
– \(C\) is at \((3, 3\sqrt{3})\).
The midpoint \(M\) of a line segment joining points \((x_1, y_1)\) and \((x_2, y_2)\) has coordinates:
\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right).
\]
Substitute the coordinates of \(A\) and \(C\):
\[
x_M = \frac{6 + 3}{2} = \frac{9}{2} = 4.5,
\]
\[
y_M = \frac{0 + 3\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}.
\]
Thus, the midpoint \(M\) of \([AC]\) has coordinates:
\[
\left( 4.5, \frac{3\sqrt{3}}{2} \right).
\]
—
Part (a)(ii): Point M lies on OB, the slope of line OB can be found by Coordinates of point O(0,0) and coordinates of M = \[
\left( 4.5, \frac{3\sqrt{3}}{2} \right).
\]
\[
\text{slope of } OM = \frac{y_M – y_O}{x_M – x_O} = \frac{\frac{3\sqrt{3}}{2} – 0}{4.5 – 0} = \frac{\frac{3\sqrt{3}}{2}}{4.5} = \frac{3\sqrt{3}}{2} \cdot \frac{1}{4.5} = \frac{3\sqrt{3}}{2} \cdot \frac{2}{9} = \frac{3\sqrt{3}}{9} = \frac{\sqrt{3}}{3}.
\]
\[
\text{slope of } OB = \frac{y_M – y_O}{x_M – x_O} = \frac{\frac{3\sqrt{3}}{2} – 0}{4.5 – 0} = \frac{\frac{3\sqrt{3}}{2}}{4.5} = \frac{3\sqrt{3}}{2} \cdot \frac{1}{4.5} = \frac{3\sqrt{3}}{2} \cdot \frac{2}{9} = \frac{3\sqrt{3}}{9} = \frac{\sqrt{3}}{3}.
\]
Hence, the equation of line passing through the origin \(O(0, 0)\) and point B, its equation is of the form \(y = mx\). Substituting the slope:
\[
y = \frac{\sqrt{3}}{3} x.
\]
Part (b): the area of quadrilateral \(OABC\), given \([OA] \perp [AB]\) and OABC is symmetric about OB
area of the quadrilateral OABC
=\[2 \times \text{Area of } \triangle OAB = 2 \times \frac{1}{2} \times OA \times AB\]
Given that \( [OA] \) is perpendicular to \( [AB] \), so \( \tan\theta = \frac{1}{\sqrt{3}} = \frac{AB}{OA} = \frac{AB}{6} \). \[ \text{Therefore } AB = \frac{6}{\sqrt{3}}, \text{ Therefore area of the quadrilateral } OABC = 6 \times \frac{6}{\sqrt{3}} \] \[ \text{area of quadrilateral } OABC = \mathbf{12\sqrt{3}} \]
……………………….Markscheme…………………….
Solution:-
(b) substituting $x=6$ into their equation
so at B $y=2\sqrt{3}$
area of triangle $OAB=\frac{1}{2}\times6\times2\sqrt{3}=6\sqrt{3}$
area of quadrilateral $OABC=12\sqrt{3}$