Given
\(
A(10,4)\) and \(
B(2,12)\). Demonstrate that any point \(
P\) equidistant from \(A\) and \(B\) then its abscissa is \(2\) less than its ordinate.
Proof:
\(PA = PB\)
Solving the equation, we get:
Answer variants:
\(\sqrt{(10-x)^2+(4-y)^2}\) \(=\) \(\sqrt{(2-x)^2+(12-y)^2}\)
\(y = x - 2\)
\(x = y - 2\)
\(\sqrt{(7 - y)^2 + (1 - x)^2}\) \(=\) \(\sqrt{(3 - y)^2 + (5 - x)^2}\)
\(\sqrt{(7 - x)^2 + (1 - y)^2}\) \(=\) \(\sqrt{(3 - x)^2 + (5 - y)^2}\)