Question

Question: 1. 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 \(\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{1}}\end{array}} \right)\). Find a linear functional f and a real number d such that \(L = \left( {f:d} \right)\).

Short answer

The linear functional is \(f\left( {x,y} \right) = 3x + 4y\), and a real number \(d = 13\) such that \(L = \left( {f:d} \right)\).

Step-by-step solution

Find the line equation

The given two points are \(\left( {\begin{array}{*{20}{c}}{ - 1}\\4\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}3\\1\end{array}} \right)\). Take \({x_1} = - 1,{\rm{ }}{y_1} = 4\), and \({x_2} = 3,{\rm{ }}{y_2} = 1\).

The line equation is given by,

\(\begin{array}{c}\frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{x - {x_1}}}{{{x_2} - {x_1}}}\\\frac{{y - 4}}{{1 - 4}} = \frac{{x + 1}}{{3 + 1}}\\4y - 16 = - 3x - 3\\3x + 4y = 13\end{array}\)

Find the linear functional f

From the above equation, take the linear functional as \(f\left( {x,y} \right) = 3x + 4y\).

Conclusion

The line Lcan be written as shown below:

\(\begin{array}{c}L = \left\{ {x \in {\mathbb{R}^2}:3x + 4y = 13} \right\}\\ = \left\{ {x \in {\mathbb{R}^2}:f\left( {x,y} \right) = 13} \right\}\\L = \left( {f:13} \right)\end{array}\)

Thus, the linear function is \(f\left( {x,y} \right) = 3x + 4y\), and a real number \(d = 13\) such that \(L = \left( {f:d} \right)\).