Step 1: Recognise that the required point lies on the perpendicular bisector of the two given points.
Any point equidistant from two fixed points $A(6, 4)$ and $B(5, 2)$ must lie somewhere on the perpendicular bisector of segment AB, since the perpendicular bisector is exactly the set of all points equidistant from A and B.
So we can find the required point by intersecting this perpendicular bisector with the given line $x + y = 5$, instead of setting up and expanding distance equations directly.
Step 2: Find the midpoint of AB.
\[ M = \left(\frac{6+5}{2}, \frac{4+2}{2}\right) = \left(\frac{11}{2}, 3\right) \]
Step 3: Find the slope of AB, then the slope of the perpendicular bisector.
\[ \text{slope of } AB = \frac{2 - 4}{5 - 6} = \frac{-2}{-1} = 2 \]
The perpendicular bisector is perpendicular to AB, so its slope is the negative reciprocal of $2$:
\[ \text{slope of perpendicular bisector} = -\frac{1}{2} \]
Step 4: Write the equation of the perpendicular bisector.
Using point slope form through $M\left(\frac{11}{2}, 3\right)$ with slope $-\frac{1}{2}$:
\[ y - 3 = -\frac{1}{2}\left(x - \frac{11}{2}\right) \]
Multiply both sides by $2$:
\[ 2y - 6 = -x + \frac{11}{2} \]
Multiply both sides by $2$ again to clear the remaining fraction:
\[ 4y - 12 = -2x + 11 \]
\[ 2x + 4y = 23 \quad \text{(equation of the perpendicular bisector)} \]
Step 5: Solve this equation together with the given line x + y = 5.
From $x + y = 5$, we get $x = 5 - y$. Substitute into $2x + 4y = 23$:
\[ 2(5 - y) + 4y = 23 \]
\[ 10 - 2y + 4y = 23 \]
\[ 10 + 2y = 23 \]
\[ 2y = 13 \]
\[ y = \frac{13}{2} \]
Step 6: Find x using x = 5 - y.
\[ x = 5 - \frac{13}{2} = \frac{10 - 13}{2} = -\frac{3}{2} \]
Final Answer:
The required point is
\[ \boxed{\left(-\frac{3}{2}, \frac{13}{2}\right)} \]