The excentre of the triangle formed by the points (1, 2), (1, 5), (5, 2) which is opposite to (1, 2) is

Now vertices of triangle are

 $$\begin{gathered} \mathrm{let} \mathrm{A}=(1, 2), \mathrm{B}=(1, 5), \mathrm{C}=(5, 2) \\ \mathrm{Now} \mathrm{a}=\mathrm{BC}=\sqrt{16+9}=5 \\ \mathrm{b}=\mathrm{AC}=\sqrt{16+0}=4 \\ \mathrm{c}=\mathrm{AB}=\sqrt{0+9}=3 \end{gathered}$$

Now excentre opposite to vertex A

$$\begin{gathered} =\left[\frac{-{\mathrm{ax}}_1+{\mathrm{bx}}_2+{\mathrm{cx}}_3}{-\mathrm{a}+\mathrm{b}+\mathrm{c}}, \frac{-{\mathrm{ay}}_1+{\mathrm{by}}_2+{\mathrm{cy}}_3}{-\mathrm{a}+\mathrm{b}+\mathrm{c}}\right] \\ =\left[\frac{-5\left(1\right)+4\left(1\right)+3\left(5\right)}{-5+4+3}, \frac{-5\left(2\right)+4\left(5\right)+3\left(2\right)}{-5+4+3}\right] \\ =\left[\frac{14}{2}, \frac{16}{2}\right] \\ =\left[7, 8\right] \end{gathered}$$