Question

If $(0,0),(a,2)$ and $(2,b)$ are the vertices of an equilateral triangle such that both $a$ and $b$ lie in $(0,2),$ then the value of $4(a+b)-ab$ is

• A. $4$
• B. $-4$
• C. $2\sqrt{3}$
• D. $4\sqrt{3}$

Answer 1

The correct option is A $4$

Let the given points be $A(0,0),B(a,2),C(2,b)$
Since, $A,B,C$ are the vertices of equilateral triangle,
$\Rightarrow AB=BC=AC$

By using distance formula, we have
$AB=\sqrt{a^2+4}\cdots \left(1\right)$

Similarly, $BC=\sqrt{(a-2{)}^2+(b-2{)}^2}\cdots \left(2\right)$
and $AC=\sqrt{4+b^2}\cdots \left(3\right)$

Equating equation $\left(1\right)$ and $\left(3\right)$, we get
$\sqrt{4+a^2}=\sqrt{4+b^2}$
$\Rightarrow a^2+4=b^2+4$
$\Rightarrow a=\pm b$
Since $a$ and $b$ lie in $(0,2),$
$\Rightarrow$ Both $a$ and $b$ are positive.
$\therefore a=b$

Now, $4(a+b)-ab=8a-a^2$

Equating equation $\left(1\right)$ and $\left(2\right)$, we get
$\sqrt{(a-2{)}^2+(b-2{)}^2}=\sqrt{a^2+4}\Rightarrow a^2-8a+4=0$

Hence, $4(a+b)-ab=8a-a^2=4$