Given:
Focuses are \[\left( \mathbf{2},\text{ }\mathbf{5} \right)\] and \[\left( -\text{ }\mathbf{3},\text{ }\mathbf{6} \right).\]
We realize that slant, \[\mathbf{m}\text{ }=\text{ }\left( \mathbf{y2}\text{ }\text{ }\mathbf{y1} \right)/\left( \mathbf{x2}\text{ }\text{ }\mathbf{x1} \right)\]
\[=\text{ }\left( \mathbf{6}\text{ }\text{ }\mathbf{5} \right)/\left( -\text{ }\mathbf{3}\text{ }\text{ }\mathbf{2} \right)\]
\[=\text{ }\mathbf{1}/\text{ }-\text{ }\mathbf{5}\text{ }=\text{ }-\text{ }\mathbf{1}/\mathbf{5}\]
We realize that two non-vertical lines are opposite to one another if and provided that their slants are negative reciprocals of one another.
Then, at that point, \[\mathbf{m}\text{ }=\text{ }\left( -\text{ }\mathbf{1}/\mathbf{m} \right)\]
\[=\text{ }-\text{ }\mathbf{1}/\left( -\text{ }\mathbf{1}/\mathbf{5} \right)\]
\[=\text{ }\mathbf{5}\]
We realize that the point (x, y) lies on the line with slant m through the proper point (x0, y0), if and provided that, its directions fulfill the condition \[\mathbf{y}\text{ }\text{ }\mathbf{y0}\text{ }=\text{ }\mathbf{m}\text{ }\left( \mathbf{x}\text{ }\text{ }\mathbf{x0} \right)\]
Then, at that point, \[\mathbf{y}\text{ }\text{ }\mathbf{5}\text{ }=\text{ }\mathbf{5}\left( \mathbf{x}\text{ }\text{ }\left( -\text{ }\mathbf{3} \right) \right)\]
\[\mathbf{y}\text{ }\text{ }\mathbf{5}\text{ }=\text{ }\mathbf{5x}\text{ }+\text{ }\mathbf{15}\]
\[\mathbf{5x}\text{ }+\text{ }\mathbf{15}\text{ }\text{ }\mathbf{y}\text{ }+\text{ }\mathbf{5}\text{ }=\text{ }\mathbf{0}\]
\[\mathbf{5x}\text{ }\text{ }\mathbf{y}\text{ }+\text{ }\mathbf{20}\text{ }=\text{ }\mathbf{0}\]
∴ The condition of the line is \[\mathbf{5x}\text{ }\text{ }\mathbf{y}\text{ }+\text{ }\mathbf{20}\text{ }=\text{ }\mathbf{0}\]