Question

If the point (0, 0) 2 Square root 3 And a, b Be the vertices of an equilateral triangle Then find a, b

Answer 1

Two vertices of an equilateral triangle are (0, 0) and (3, √3). Let the third vertex of the equilaterla triangle be (x, y) Distance between (0, 0) and (x, y) = Distance between (0, 0) and (3, √3)  = Distance between (x, y) and (3, √3) √(x2 + y2) =√(32 + 3)  = √[(x - 3)2 + (y - √3)2] x2 + y2 = 12 x2 + 9 - 6x + y2 + 3 - 2√3y  = 12 24 -  6x - 2√3y  = 12 - 6x - 2√3y  = - 12 3x + √3y = 6 x  = (6 - √3y) / 3 ⇒ [(6 - √3y)/3]2 + y2  = 12  ⇒ (36 + 3y2 - 12√3y) / 9 + y2 = 12 ⇒ 36 + 3y2 - 12√3y + 9y2 = 108  ⇒ - 12√3y + 12y2 - 72 = 0  ⇒ -√3y + y2 - 6 = 0  ⇒ (y - 2√3)(y + √3) = 0  ⇒ y = 2√3 or - √3 If y = 2√3, x = (6 - 6) / 3 = 0 If y = -√3, x = (6 + 3) / 3 = 3 So, the third vertex of the equilateral triangle = (0, 2√3) or (3, -√3).

Answer 2

Coordinates of C is

$C\left ( \left ( \sqrt{3}-3 \right )/2,\left ( \sqrt{3}+3 \right )/2 \right )$   or $C\left ( \left ( \sqrt{3}+3 \right )/2,\left ( \sqrt{3}-3 \right )/2 \right )$

Step-by-step explanation:

Given,

Let ABC is an equilateral triangle

Coordinates of   A                 $A\left ( \sqrt{3},\sqrt{3} \right )$

Coordinates of B                    $B\left ( 0,0 \right )$

Coordinates of C                   $C\left ( a,b \right )$

Find      ( a,b)

Solution,

Distance between two point formula $=\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$

For equilateral triangle $AB=AC=BC$

$\sqrt{\left ( \sqrt{3} -0\right )^{2}+\left ( \sqrt{3}-0 \right )^{2}}=\sqrt{\left ( \sqrt{3}-a \right )^{2}+\left ( \sqrt{3}-b \right )^{2}}=\sqrt{\left ( 0-a \right )^{2}+\left ( 0-b \right )^{2}}$

$(AB)^{2}=\left ( \sqrt{3} \right )^{2}+\left ( \sqrt{3} \right )^{2}=3+3=6$

$a^{2}+b^{2}=\left ( \sqrt{3}-a \right )^{2}+\left ( \sqrt{3}-b \right )^{2}$

$=a^{2}+b^{2}=3+a^{2}-2\sqrt{3}a+3+b^{2}-2\sqrt{3}b$

$=2\sqrt{3}a+2\sqrt{3}b=6$

$=a+b=\sqrt{3}$       (1)

$a^{2}+b^{2}=6$        side square

$\left ( a+b \right )^{2}=\left ( \sqrt{3} \right )^{2}$

$a^{2}+b^{2}+2ab=3$

$6+2ab=3$

$2ab=-3$

$ab=-3/2$

$a+b=\sqrt{3}$

$a=\sqrt{3}-b$

$\left ( \sqrt{3}-b \right )b=-3/2$

$\sqrt{3}b-b^{2}=-3/2$

$2b^{2}-2\sqrt{3}b=3$

$2b^{2}-2\sqrt{3}b-3=0$

$b= \left ( 2\sqrt{3} \pm\sqrt{\left ( \left ( 2\sqrt{3} \right )^{2}-4\times 2\times- 3 \right )} \right )/2\times 2$

$b=(2\sqrt{3} \pm 6)/(4)$

$b=(\sqrt{3} \pm 3)/2$

$b=\left ( \sqrt{3}+3 \right )/2$    or

$b=\left ( \sqrt{3}-3 \right )/2$

$a+b=\sqrt{3}$

$a=\left ( \sqrt{3}-3 \right )/2$   or

$a=\left ( \sqrt{3}+3 \right )/2$

Coordinates of C are

$C\left ( \left ( \sqrt{3}-3 \right )/2,\left ( \sqrt{3}+3 \right )/2 \right )$

or

$C\left ( \left ( \sqrt{3}+3 \right )/2,\left ( \sqrt{3}-3 \right )/2 \right )$