Question: 2. Let Lbe the line in \({\mathbb{R}^{\bf{2}}}\) through the points \(\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{4}}\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{ - {\bf{1}}}\end{array}} \right)\). Find a linear functional f and a real number d such that \(L = \left( {f:d} \right)\).

The given two points are \(\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}{ - 2}\\{ - 1}\end{array}} \right)\). Take \({x_1} = 1,{y_1} = 4\), and \({x_2} = - 2,{y_2} = - 1\).

The line equation is given by,

\(\begin{array}{c}\frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{x - {x_1}}}{{{x_2} - {x_1}}}\\\frac{{y - 4}}{{ - 1 - 4}} = \frac{{x - 1}}{{ - 2 - 1}}\\ - 3\left( {y - 4} \right) = - 5\left( {x - 1} \right)\\ - 3y + 12 = - 5x + 5\\5x - 3y = - 7\end{array}\)

From the above equation, take the linear functionalas \(f\left( {x,y} \right) = 5x - 3y\).

The line L is as follows:

\(\begin{array}{c}L = \left\{ {x \in {\mathbb{R}^2}:5x - 3y = - 7} \right\}\\ = \left\{ {x \in {\mathbb{R}^2}:f\left( {x,y} \right) = - 7} \right\}\\L = \left( {f: - 7} \right)\end{array}\)

Therefore, linear functional is \(f\left( {x,y} \right) = 5x - 3y\), and a real number \(d = - 7\) such that \(L = \left( {f:d} \right)\).