If (-4,3) & (4,3) are two …

Let B(-4, 3) and C(4, 3) be the given two vertices of an equilateral triangle.
Let A(x, y) be the third vertex.
Then, we have
AB = BC = AC
Consider AB = BC
{tex}\Rightarrow {/tex} AB² = BC²
{tex}\Rightarrow{/tex} (-4 - x)² + (3 - y)² = (4 + 4)² + (3 - 3)²
{tex}\Rightarrow{/tex}16 + x² + 8x + 9 + y² - 6y = 64
{tex}\Rightarrow{/tex} x² + y² + 8x - 6y = 39 .....(i)
Consider AB = AC
{tex}\Rightarrow{/tex} AB² = AC²
{tex}\Rightarrow{/tex} (-4 - x)² + (3 - y)² = (4 - x)² + (3 - y)²
{tex}\Rightarrow{/tex} 16 + x² + 8x = 16 + x² - 8x
{tex}\Rightarrow{/tex} 16x = 0
{tex}\Rightarrow{/tex} x = 0
Consider BC = AC
{tex}\Rightarrow{/tex} BC² = AC²
{tex}\Rightarrow{/tex} (4 + 4)² + (3 - 3)² = (4 - x)² + (3 - y)²
{tex}\Rightarrow{/tex} 8² + 0 = (4 - 0)² + (3 - y)²
{tex}\Rightarrow{/tex} 64 = 16 + (3 - y)²
(3 - y)² = 48
3 - y = {tex}\pm 4 \sqrt { 3 }{/tex}
y = 3 {tex} \pm 4 \sqrt { 3 }{/tex}
Thus, the coordinates of third vertex when origin lies in the interior of a triangle = {tex}( 0,3 - 4 \sqrt { 3 } ){/tex}