Given:

The right bisector of a line section divides the line fragment at \[\mathbf{90}{}^\circ .\]

End-points of the line portion AB are given as \[\mathbf{A}\text{ }\left( \mathbf{3},\text{ }\mathbf{4} \right)\text{ }\mathbf{and}\text{ }\mathbf{B}\text{ }\left( \text{ }\mathbf{1},\text{ }\mathbf{2} \right).\]

Leave mid-point of AB alone (x, y)

\[\mathbf{x}\text{ }=\text{ }\left( \mathbf{3}-\mathbf{1} \right)/\mathbf{2}=\text{ }\mathbf{2}/\mathbf{2}\text{ }=\text{ }\mathbf{1}\]

\[\mathbf{y}\text{ }=\text{ }\left( \mathbf{4}+\mathbf{2} \right)/\mathbf{2}=\text{ }\mathbf{6}/\mathbf{2}\text{ }=\text{ }\mathbf{3}\]

\[\left( \mathbf{x},\text{ }\mathbf{y} \right)\text{ }=\text{ }\left( \mathbf{1},\text{ }\mathbf{3} \right)\]

Leave the incline of line AB alone m1

\[\mathbf{m1}\text{ }=\text{ }\left( \mathbf{2}\text{ }\text{ }\mathbf{4} \right)/\left( -\text{ }\mathbf{1}\text{ }\text{ }\mathbf{3} \right)\]

\[=\text{ }-\text{ }\mathbf{2}/\left( -\text{ }\mathbf{4} \right)\]

\[=\text{ }\mathbf{1}/\mathbf{2}\]

Also, let the incline of the line opposite to AB be m2

\[\mathbf{m2}\text{ }=\text{ }-\text{ }\mathbf{1}/\left( \mathbf{1}/\mathbf{2} \right)\]

\[=\text{ }-\text{ }\mathbf{2}\]

The condition of the line going through (1, 3) and having an incline of – 2 is

\[\left( \mathbf{y}\text{ }\text{ }\mathbf{3} \right)\text{ }=\text{ }-\text{ }\mathbf{2}\text{ }\left( \mathbf{x}\text{ }\text{ }\mathbf{1} \right)\]

\[\mathbf{y}\text{ }\text{ }\mathbf{3}\text{ }=\text{ }\text{ }\mathbf{2x}\text{ }+\text{ }\mathbf{2}\]

\[\mathbf{2x}\text{ }+\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{5}\]

∴ The necessary condition of the line is \[\mathbf{2x}\text{ }+\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{5}\]