Chapter 3.6, Problem 2CYP

To determine

Construct a line perpendicular to l through P(1,7). Find the distance from P to l .

Answer to Problem 2CYP

  $2\sqrt{5}$ units

Explanation of Solution

Given :

Line l contains points at (1,2) and (5,4) .

Formula Used:

Distance between a point and a line :

The distance between a line and a point not on the line is the length of the segment perpendicular to the line from the point.

Calculation :

Draw the line l and a line $\overleftrightarrow{PQ}$ perpendicular to it that passes through P(1,7) :

  

First , find the equations of line l and $\overleftrightarrow{PQ}$ .

By Point-Slope form of line , we find the equation of line l passing through points (1,2) and (5,4):

  $m=slope=\left(\frac{y_2-y_1}{x_2-x_1}\right)=\frac{4-2}{5-1}=\frac{2}{4}=\frac{1}{2}$

Now, the equation of line l is :

  $\begin{array}{l}\left(y-y_1\right)=m\left(x-x_1\right)\\ \Rightarrow y-2=\frac{1}{2}(x-1)\\ \Rightarrow 2y-4=x-\mathrm{1............}[multiply\text{ 2 each side]}\\ \Rightarrow -x+2y-3=\mathrm{0............}[subtract\text{ x-1 from each side]}\end{array}$

The product of slopes of two perpendicular lines is -1 .

So, slope of the line $\overleftrightarrow{PQ}$ is $m_{\perp }=\frac{-1}{m}=\frac{-1}{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}=-2$

It passes through the point (1,7) , so the equation of $\overleftrightarrow{PQ}$ by point slope form is :

  $\begin{array}{l}\left(y-y_1\right)=m\left(x-x_1\right)\\ \Rightarrow y-7=-2(x-1)\\ \Rightarrow y-7=-2x+2\\ \Rightarrow 2x+y-9=\mathrm{0............}[subtract\text{ -2x+2 from each side]}\end{array}$

Now , we find the point of intersection of line l and $\overleftrightarrow{PQ}$ :

  $-x+2y-3=0$ ........(1)

  $2x+y-9=0$ ..........(2)

Multiply equation (1) by 2 and add:

  $\begin{array}{l}\Rightarrow 4y+y-6-9=0\\ \Rightarrow 5y-15=0\\ \Rightarrow 5y=\mathrm{15........}[add\text{ 15 each side]}\\ \Rightarrow y=\mathrm{3............}[\text{divide each side by 5]}\end{array}$

Substitute in equation (1) and solve for x :

  $\begin{array}{l}-x+2y-3=0\\ \Rightarrow -x+2(3)-3=0\\ \Rightarrow -x+6-3=0\\ \Rightarrow x=3\end{array}$

So, the two lines intersect at Q(3,3)

Now, the distance from P to l is the distance between the points P(1,7) and Q(3,3) :

  $\begin{array}{l}=\sqrt{{\left(y_2-y_1\right)}^2+{\left(x_2-x_1\right)}^2}\\ =\sqrt{{\left(3-7\right)}^2+{\left(3-1\right)}^2}\\ =\sqrt{16+4}\\ =\sqrt{20}\\ =2\sqrt{5}\end{array}$