Chapter 16, Problem 11RP

To determine

To find: the equation of two lines that are parallel to the given equations and its three units from the points.

Answer to Problem 11RP

  $\begin{array}{l}y-2.8=\frac{1}{4}\left(x+1.05\right).\\ y-7.88=\frac{1}{4}\left(x-0.22\right).\end{array}$

Explanation of Solution

Given:

The equation is:

  $x-4y=7$ .

The point is:

  $\left(5,1\right)$ .

Concept used:

If lines are parallel to each other then:

  $m_1=m_2.$

The distance formula:

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Calculation:

According to the given the straight line is:

  $\begin{array}{l}x-4y=7.\\ y=\frac{1}{4}x-\frac{7}{4}.\end{array}$

Slope here is $\frac{1}{4}$ .

Since the lines are parallel to each other therefore:

  $m_1=m_2=\frac{1}{4}$ .

Let point $\left(h,k\right)$ replace the $\left(x,y\right)$ in the equation of line.

  $\begin{array}{l}h-4k=7.\\ h=7+4k.\end{array}$

Distance between a point $A$

  $\left(5,1\right)$ to the point $P$

  $\left(h,k\right)$ is $3$ units.

  $\begin{array}{l}PA=3.\\ P{A}^2=9.\\ {\left(h-5\right)}^2+{\left(k-1\right)}^2=\mathrm{9..........}\left(i\right)\end{array}$

Putting the value of $h=7+4k$ in equation $\left(i\right)$ :

  $\begin{array}{l}{\left(7+4k-5\right)}^2+{\left(k-1\right)}^2=9.\\ 2^2+{\left(4k\right)}^2+16k+k^2-2k+1=9.\\ 17k^2+14k-4=0.\end{array}$

Solving through discriminant method:

  $\begin{array}{l}k=-\frac{7}{17}-\frac{3\sqrt{3}}{17}.\\ k=\frac{3\sqrt{13}}{17}-\frac{7}{17}.\end{array}$

  $\begin{array}{l}k=-\frac{7}{17}-\frac{3\sqrt{3}}{17}=-1.05\\ k=\frac{3\sqrt{13}}{17}-\frac{7}{17}=0.22\end{array}$

From the value of $k$ the value of $h$ can be measured.

  $\begin{array}{l}h=7+4\left(-1.05\right)=\mathrm{2.8.}\\ h=7+4\left(0.22\right)=7.88\end{array}$

Required point are:

  $\left(-1.05,2.8\right)\text{ and }\left(0.22,7.88\right)$ .

Since, slope $m_1=m_2=\frac{1}{4}$ .

Hence, the equation of two lines will be:

  $\begin{array}{l}y-2.8=\frac{1}{4}\left(x+1.05\right).\\ y-7.88=\frac{1}{4}\left(x-0.22\right).\end{array}$