Exercise 2.1.9

To find the slope \(m_1\) of \(L_1\) we solve \(2x+3y=6\) for \(y\), obtaining the result \(y=-\frac{2}{3}x+2\) and conclude that \(m_1=-\frac{2}{3}\).

Since \(L_2\) is parallel to \(L_1\) it has the same slope, so \(m_2=\frac{2}{3}\) also.

Since the line \(L_2\) contains the point \((2,-1)\) we can use the point-slope formula to get \(y+1=\left(-\frac{2}{3}\right)(x-2)\).

Solving for \(y\) gives \(y=-\frac{2}{3}x+\frac{1}{3}\). Clearing fractions by multiplying both sides by \(3\) gives \(3y=-2x+1\) from which we obtain the general equation \(2x+3y=1\). Compare this to the equation of line \(L_1\): \(2x+3y=6\). All lines parallel to the line \(2x+3y=6\) will be of the form \(2x+3y=C\). Notice that if we substitute the coordinates of the point \((2,-1)\) into the expression \(2x+3y\) we obtain the number \(C=2(2)+3(-1)=1\). Does this suggest a shorter way to work such a problem?

Line \(L_3\) is perpendicular to line \(L_1\) so its slope is \(m_3=-\frac{1}{m_1}=\frac{3}{2}\).

Line \(L_3\) also contains the point \((2,-1)\) so the point slope formula gives \(y+1=\left(\frac{3}{2}\right)(x-2)\).

Solving for \(y\) gives \(y=\frac{3}{2}x-4\). Clearing the fractions gives \(2y=3x-8\) resulting in the general equation \(3x-2y=8\).

Compare the equation of \(L_3:~~3x-2y=8\) to the equation of \(L_1:~~2x+3y=6\).

[Graphs of L1, L2, and L3]