Chapter 3.6, Problem 43HP

To determine

To Find: The value of $a$ and the equations of the parallel lines.

Answer to Problem 43HP

  $\begin{array}{l}{y}_n=\frac{-1}{2}x+6\\ y_m=-\frac{1}{2}x+\frac{7}{2}\end{array}$

  $\begin{array}{l}{y}_n=\frac{1}{2}x+6\\ y_m=\frac{1}{2}x+\frac{7}{2}\end{array}$

Explanation of Solution

Given:Suppose a line perpendicular to the pair of parallel lines intersect the lines at the point $(a,4)and(0,6)$ and the distance between the parallel lines is $\sqrt{5}.$

Given: Points= $(a,4)and(0,6)$

The distance between the lines is $\sqrt{5}.$

Thus

  $\begin{array}{l}\sqrt{5}=\sqrt{(0-a)+{(4-6)}^2}\\ a^2+4={\left(\sqrt{5}\right)}^2\\ a^2+4=5\\ a^2=1\\ a=\pm 1\end{array}$

Finding the slope of the line.

  $slope=\frac{6-4}{0-a}=-\frac{2}{a}$

Now the lines are perpendicular.

Therefore

  $\begin{array}{l}{m}_1\times m_2=-1\\ m_2=\frac{-1}{\frac{-2}{a}}\\ m_2=\frac{a}{2}\end{array}$

Thus

Equation of the line $m.$

  $\begin{array}{l}y-y_1=m(x-x_1)\\ y-4=\frac{a}{2}(x-a)\\ y=\frac{a}{2}x-\frac{a^2}{2}+4\end{array}$

For line $n$

Slope

  $m_n=\frac{-1}{\frac{-2}{a}}=\frac{a}{2}$

Equation

  $\begin{array}{l}y-y_1=m(x-x_1)\\ y-6=\frac{a}{2}(x-0)\\ y=\frac{a}{2}x+6\end{array}$

Now when $a=-1$

  $\begin{array}{l}y=\frac{a}{2}x-\frac{a^2}{2}+4\\ y_m=-\frac{1}{2}x+\frac{1^2}{2}+4\\ y_m=-\frac{1}{2}x+\frac{7}{2}\\ y_n=\frac{a}{2}x+6\\ y_n=\frac{-1}{2}x+6\end{array}$

Now

When $a=1$

  $\begin{array}{l}y=\frac{a}{2}x-\frac{a^2}{2}+4\\ y_m=\frac{1}{2}x-\frac{1^2}{2}+4\\ y_m=\frac{1}{2}x+\frac{7}{2}\\ y_n=\frac{a}{2}x+6\\ y_n=\frac{1}{2}x+6\end{array}$