Find the coordinates of the point …

We have
A {tex}\rightarrow{/tex} (5, 1)
B {tex}\rightarrow{/tex} (-3, -7)
C {tex}\rightarrow{/tex} (7, -1)
Let the point P(x, y) be equidistant from the three given points A, B, and C.
Then PA = PB = PC
{tex}\Rightarrow{/tex} PA² = PB² = PC²
First two give
PA² = PB²
{tex}\Rightarrow{/tex} (x - 5)² + (y - 1)² = (x + 3)² + (y + 7)²
{tex}\Rightarrow{/tex} x² - 10x + 25 + y² - 2y + 1 = x² + 6x + 9 + y² + 14y + 49
{tex}\Rightarrow{/tex} 16x + 16y + 32 = 0
{tex}\Rightarrow{/tex} x + y + 2 = 0 ....Dividing throughout by 16 ....(1)
Last two give
PB² = PC²
{tex}\Rightarrow{/tex} (x + 3)² + (y + 7)² = (x - 7)² + (y + 1)²
{tex}\Rightarrow{/tex} x² + 6x + 9 + y² + 14y + 49 = x² - 14x + 49 + y² + 2y + 1
{tex}\Rightarrow{/tex} 20x + 12y + 8 = 0
{tex}\Rightarrow{/tex} 5x + 3y + 2 = 0 ....Dividing throughout by 4 ....(2)
Multiplying equation (1) by 3, we get
3x + 3y + 6 = 0 ....(3)
Subtracting equation (3) from equation (2), we get
2x - 4 = 0
{tex}\Rightarrow{/tex} 2x = 4
{tex}\Rightarrow x = \frac{4}{2} = 2{/tex}
Subtracting x = 2 in equation (1), we get
2 + y + 2 = 0
{tex}\Rightarrow{/tex} y + 4 = 0
{tex}\Rightarrow{/tex} y = -4
Hence, the required points is (2, -4).