Q2 As shown in the figure, on an $xy$-plane whose origin is O, let us consider an isosceles triangle ABC satisfying $\mathrm { AB } = \mathrm { AC }$. Furthermore, suppose that side AB passes through $\mathrm { P } ( - 1,5 )$ and side AC passes through $\mathrm{Q}(3, 3)$.
Let us consider the radius of the inscribed circle of the triangle ABC.
Denote the straight line passing through the two points A and B by $\ell _ { 1 }$ and the straight line passing through the two points A and C by $\ell _ { 2 }$. When we denote the slope of $\ell _ { 1 }$ by $a$, the equations of $\ell _ { 1 }$ and $\ell _ { 2 }$ are
$$\begin{aligned} & \ell _ { 1 } : y = a x + a + \mathbf { M } , \\ & \ell _ { 2 } : y = - a x + \mathbf { N } a + \mathbf { O } . \end{aligned}$$
Denote the center and the radius of the inscribed circle by I and $r$, respectively. Then the coordinates of I are $\left( \mathbf { P } - \frac { \mathbf { Q } } { a } , r \right)$.
Hence $r$ can be expressed in terms of $a$ as
$$r = \frac { \mathbf { R } } { \mathbf { T } + \sqrt { \mathbf { S } } }$$
In particular, when $r = \frac { 5 } { 2 }$, the coordinates of vertex A are $\left( \frac { \mathbf { V } } { \mathbf{U} } , \frac { \mathbf { X Y } } { \mathbf { W } } \right)$.