Question
Let $B(1, - 3)$ and $D(0,4)$ represent two vertices of rhombus $ABCD$ in ( $x,y$ ) plane, then coordinates of vertex $A$ if $\angle BAD = 60^{\circ}$ can be equal to:
Options
Solution
Sol. [Key Concept: With $B(1, - 3),D(0,4)$ and $\angle\ BAD{= 60}^{\circ}$, $A$ has two positions symmetric about diagonal $BD$.]
$${tan\theta = \frac{1}{7},sin\theta = \frac{1}{5\sqrt{2}},cos = \frac{7}{5\sqrt{2}} }{AP \equiv \frac{5}{2}\sqrt{2}cot30^{\circ} = \frac{5}{2}\sqrt{2}\sqrt{3} }{A \equiv \left( \frac{1}{2} - \frac{5\sqrt{2}\sqrt{3}}{2}\frac{7}{5\sqrt{2}},\frac{1}{2} - \frac{5\sqrt{2}\sqrt{3}}{2}\frac{1}{5\sqrt{2}} \right) }$$or, $\left( \frac{1}{2} + \frac{5\sqrt{2}\sqrt{3}}{2}\frac{7}{5\sqrt{2}},\frac{1}{2} + \frac{5\sqrt{2}\sqrt{3}}{2}\frac{1}{5\sqrt{2}} \right)$
$$\Rightarrow A = \left( \frac{1 - 7\sqrt{3}}{2},\frac{1 - \sqrt{3}}{2} \right) $$or $\left( \frac{1 + 7\sqrt{3}}{2},\frac{1 + \sqrt{3}}{2} \right)$
✔ (a) $(\frac{1 - 7\sqrt{3}}{2},\frac{1 - \sqrt{3}}{2})$ is one position of $A$.
✔ (b) $(\frac{1 + 7\sqrt{3}}{2},\frac{1 + \sqrt{3}}{2})$ is the other position.
✘ (c) wrong sign combination.
✘ (d) wrong sign combination.
Create a free account to view solution
View Solution Free