Question

If two vertices of an equilateral triangle be $(0,0),(3,\sqrt{3})$, find the third vertex. [4 MARKS]

Answer 1

Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark

O(0 , 0) and $A(3,\sqrt{3})$ be the points and let B ( x, y) be the third vertex of equilateral $\Delta OAB$

Then,

OA = OB = AB

$\Rightarrow {OA}^2={OB}^2={AB}^2$

We have ${OA}^2={(3-0)}^2+{\left(\sqrt{3}\right)-0)}^2=12$,

${OB}^2=x^2+y^2$

and, ${AB}^2={(x-3)}^2+{y-\sqrt{3})}^2$

$\Rightarrow {AB}^2=x^2+y^2-6x-2\sqrt{3}y+12$

$\therefore {OA}^2={OB}^2={AB}^2$

$\Rightarrow {OA}^2={OB}^2$ and ${OB}^2={AB}^2$

$\Rightarrow x^2+y^2=12$

and, $x^2+y^2=x^2+y^2-6x-2\sqrt{3}y+12$

$\Rightarrow x^2+y^2=12and6x+2\sqrt{3}y=12$

$\Rightarrow x^2+y^2=12and3x+\sqrt{3}y=6$

$\Rightarrow x^2+{\left(\frac{6-3x}{\sqrt{3})}\right)}^2=12$

$[\because 3x+\sqrt{3}y=6\therefore y=\frac{6-3x}{\sqrt{3}}]$

We have,

$\Rightarrow 3x^2+{(6-3x)}^2=36$

$\Rightarrow 12x^2-36x=0$

$\Rightarrow x=0,3$

$\therefore x=0\Rightarrow \left(\sqrt{3}\right)y=6$

$\Rightarrow y=\frac{6}{\sqrt{3}}=2\sqrt{3}$ [Putting x =0 in $(3x+\sqrt{3}y)=6$ ]

and , $x=3\Rightarrow 9+\left(\sqrt{3}\right)y=6$

$\Rightarrow y=\frac{6-9}{\sqrt{3}}=-\sqrt{3}$ [Putting x = 3 in $3x+\left(\sqrt{3}y\right)=6$]

Hence . the coordinates of the third vertex b are $(0,2\sqrt{3})$ or $(3,-\sqrt{3})$