We realize that condition of the line making blocks an and b on x-and y-axis, individually, is \[\mathbf{x}/\mathbf{a}\text{ }+\text{ }\mathbf{y}/\mathbf{b}\text{ }=\text{ }\mathbf{1}\text{ }.\text{ }\ldots \text{ }\left( \mathbf{1} \right)\]
Given: amount of captures = 9
\[\mathbf{a}\text{ }+\text{ }\mathbf{b}\text{ }=\text{ }\mathbf{9}\]
\[\mathbf{b}\text{ }=\text{ }\mathbf{9}\text{ }\text{ }\mathbf{a}\]
Presently, substitute worth of b in the above condition, we get
\[\mathbf{x}/\mathbf{a}\text{ }+\text{ }\mathbf{y}/\left( \mathbf{9}\text{ }\text{ }\mathbf{a} \right)\text{ }=\text{ }\mathbf{1}\]
Given: the line goes through the point (2, 2),
In this way, \[\mathbf{2}/\mathbf{a}\text{ }+\text{ }\mathbf{2}/\left( \mathbf{9}\text{ }\text{ }\mathbf{a} \right)\text{ }=\text{ }\mathbf{1}\]
\[\left[ \mathbf{2}\left( \mathbf{9}\text{ }\text{ }\mathbf{a} \right)\text{ }+\text{ }\mathbf{2a} \right]/\mathbf{a}\left( \mathbf{9}\text{ }\text{ }\mathbf{a} \right)\text{ }=\text{ }\mathbf{1}\]
\[\left[ \mathbf{18}\text{ }\text{ }\mathbf{2a}\text{ }+\text{ }\mathbf{2a} \right]/\mathbf{a}\left( \mathbf{9}\text{ }\text{ }\mathbf{a} \right)\text{ }=\text{ }\mathbf{1}\]
\[\mathbf{18}/\mathbf{a}\left( \mathbf{9}\text{ }\text{ }\mathbf{a} \right)\text{ }=\text{ }\mathbf{1}\]
\[\mathbf{18}\text{ }=\text{ }\mathbf{a}\text{ }\left( \mathbf{9}\text{ }\text{ }\mathbf{a} \right)\]
\[\mathbf{18}\text{ }=\text{ }\mathbf{9a}\text{ }\text{ }\mathbf{a2}\]
\[\mathbf{a2}\text{ }\text{ }\mathbf{9a}\text{ }+\text{ }\mathbf{18}\text{ }=\text{ }\mathbf{0}\]
After factorizing, we get
\[\mathbf{a2}\text{ }\text{ }\mathbf{3a}\text{ }\text{ }\mathbf{6a}\text{ }+\text{ }\mathbf{18}\text{ }=\text{ }\mathbf{0}\]
\[\left( \mathbf{a}\text{ }\text{ }\mathbf{3} \right)\text{ }\text{ }\mathbf{6}\text{ }\left( \mathbf{a}\text{ }\text{ }\mathbf{3} \right)\text{ }=\text{ }\mathbf{0}\]
\[\left( \mathbf{a}\text{ }\text{ }\mathbf{3} \right)\text{ }\left( \mathbf{a}\text{ }\text{ }\mathbf{6} \right)\text{ }=\text{ }\mathbf{0}\]
\[\mathbf{a}\text{ }=\text{ }\mathbf{3}\text{ }\mathbf{or}\text{ }\mathbf{a}\text{ }=\text{ }\mathbf{6}\]
Allow us to substitute in (1),
\[\mathbf{Case}\text{ }\mathbf{1}\text{ }\left( \mathbf{a}\text{ }=\text{ }\mathbf{3} \right):\]
Then, at that point, \[\mathbf{b}\text{ }=\text{ }\mathbf{9}\text{ }\text{ }\mathbf{3}\text{ }=\text{ }\mathbf{6}\]
\[\mathbf{x}/\mathbf{3}\text{ }+\text{ }\mathbf{y}/\mathbf{6}\text{ }=\text{ }\mathbf{1}\]
\[\mathbf{2x}\text{ }+\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{6}\]
\[\mathbf{2x}\text{ }+\text{ }\mathbf{y}\text{ }\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}\]
\[\mathbf{Case}\text{ }\mathbf{2}\text{ }\left( \mathbf{a}\text{ }=\text{ }\mathbf{6} \right):\]
Then, at that point, \[\mathbf{b}\text{ }=\text{ }\mathbf{9}\text{ }\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{3}\]
\[\mathbf{x}/\mathbf{6}\text{ }+\text{ }\mathbf{y}/\mathbf{3}\text{ }=\text{ }\mathbf{1}\]
\[\mathbf{x}\text{ }+\text{ }\mathbf{2y}\text{ }=\text{ }\mathbf{6}\]
\[\mathbf{x}\text{ }+\text{ }\mathbf{2y}\text{ }\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}\]
∴ The condition of the line is \[\mathbf{2x}\text{ }+\text{ }\mathbf{y}\text{ }\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}\text{ }\mathbf{or}\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{2y}\text{ }\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}.\]