Prove that the point $P(7, 5)$ is equidistant from the points $A(2, 4)$ and $B(6, 10)$.

(N/A) To prove that point $P(7, 5)$ is equidistant from $A(2, 4)$ and $B(6, 10)$,we need to show that the distance $PA$ is equal to the distance $PB$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$1$. Distance $PA = \sqrt{(7 - 2)^2 + (5 - 4)^2} = \sqrt{5^2 + 1^2} = \sqrt{25 + 1} = \sqrt{26}$.
$2$. Distance $PB = \sqrt{(7 - 6)^2 + (5 - 10)^2} = \sqrt{1^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26}$.
Since $PA = PB = \sqrt{26}$,the point $P(7, 5)$ is equidistant from points $A(2, 4)$ and $B(6, 10)$.