Chapter 3, Problem 21PT

To determine

The distance between P to l.

Answer to Problem 21PT

The required distance is 2.68 units.

Explanation of Solution

Given:

Line l contains points: $\left(6,5\right)\text{and}\left(2,3\right)$

And

Point: $P\left(2,6\right)$

Solution:

Slope of line l:

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{3-5}{2-6}\\ =\frac{1}{2}\end{array}$

Now,

Using point,

  $\left(6,5\right)$

Equation of line:

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-5=\frac{1}{2}\left(x-6\right)\\ y=\frac{1}{2}x-3+5\\ y=\frac{1}{2}x+2\mathrm{.......................}\left(1\right)\end{array}$

Now,

Suppose a line “s” perpendicular to given line l passing through point $P\left(2,6\right)$ .

So, the slope of line “s” will be negative reciprocal of the slope of given line l which is 1/2.

Therefore,

Slope of line “s” is -2.

Equation of line “s”:

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-6=-2\left(x-2\right)\\ y=-2x+4+6\\ y=-2x+10\mathrm{.......................}\left(2\right)\end{array}$

Now, on solving equation 1 and 2,

Point of intersection:

  $\begin{array}{l}x=\frac{16}{5}\\ x=3.2\end{array}$

So,

On putting the value in equation 2,

  $\begin{array}{l}y=\frac{18}{5}\\ y=3.6\end{array}$

Point of intersection: $\left(3.2,3.6\right)$

Suppose the point of intersection is T $\left(3.2,3.6\right)$ .

Therefore,

The distance between the point $P\left(2,6\right)$ and T $\left(3.2,3.6\right)$ ,

Using distance formula:

  $\begin{array}{l}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ d=\sqrt{{\left(3.2-2\right)}^2+{\left(3.6-6\right)}^2}\\ d=\sqrt{7.2}\\ d=2.68\end{array}$

Thus, the required distance is 2.68 units.