Straight LineHardBloom L3
Question
If one vertex of an equilateral triangle of side ' $a$ ' lies at origin and the other lies on the line $x - \sqrt{3}y = 0$, then the coordinates of the third vertex are:
Options
A.$(0,a)$
B.$\left( \frac{\sqrt{3}a}{2}, - \frac{a}{2} \right)$
C.$(0, - a)$
D.$\left( - \frac{\sqrt{3}a}{2},\frac{a}{2} \right)$
Solution
Sol. [Key Concept: With one vertex at $O$ and another on $ax - \sqrt{3}y = 0$, the third vertex has four possible positions (two lines × two sides).]
Possible ordinates of $C$ are ( 0 a) , ( $0, - a$ ),
$$\left( \frac{a\sqrt{3}}{2},\frac{- a}{2} \right),\left( \frac{- a\sqrt{3}}{2},\frac{a}{2} \right)$$
✔ (a) $(0,a)$ is one admissible third vertex.
✔ (b) $(\frac{\sqrt{3\ }a}{2}, - \frac{a}{2})$ is admissible.
✔ (c) $(0, - a)$ is admissible.
✔ (d) $( - \frac{\sqrt{3\ }a}{2},\frac{a}{2})$ is admissible.
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