61.

The vertices of a triangle are A(3, 7), B (3, 4) and C (5, 4). The equation of the bisector of the angle ABC is

• y = x + 1

• y = x - 1

• y = 3x - 5

• y = x

Equation of line AB is

$$\begin{gathered} \mathrm{y} - 7 = \frac{4 - 7}{3 - 3}\left(\mathrm{x} - 3\right) \\ \Rightarrow \left(\mathrm{x} - 3\right) = 0 ...\left(\mathrm{i}\right) \\ \mathrm{and} \mathrm{equation} \mathrm{of} \mathrm{line} \mathrm{BC} \mathrm{is} \\ \mathrm{y} - 4 = \frac{4 - 4}{5 - 3}\left(\mathrm{x} - 3\right) \\ \Rightarrow \mathrm{y} - 4 = 0 \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Equation} \mathrm{of} \mathrm{the} \mathrm{bisector} \mathrm{of} \mathrm{angle} \mathrm{ABC} \mathrm{is} \\ \frac{\mathrm{x} - 3}{\sqrt{1^2}} = \pm \left(\frac{\mathrm{y} - 4}{\sqrt{1^2}}\right) \\ \Rightarrow \mathrm{x} - 3 = \pm \left(\mathrm{y} - 4\right) \\ \Rightarrow \mathrm{x} - \mathrm{y} + 1 = 0 \\ \Rightarrow \mathrm{x} + \mathrm{y} - 7 = 0 \\ \Rightarrow \mathrm{y} = \mathrm{x} +\hspace{0.17em}1 \end{gathered}$$