Chapter 3, Problem 20PT

To determine

The distance between P to l.

Answer to Problem 20PT

The required distance is 3.5 units.

Explanation of Solution

Given:

Line l contains points: $\left(-4,2\right)\text{and}\left(3,-5\right)$

And

Point: $P\left(1,2\right)$

Solution:

Slope of line l:

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{-5-2}{3-\left(-4\right)}\\ =\frac{-7}{7}\\ =-1\end{array}$

Now,

Using point,

  $\left(-4,2\right)$

Equation of line:

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-2=-1\left(x-\left(-4\right)\right)\\ y=-x-4+2\\ y=-x-2\mathrm{.......................}\left(1\right)\end{array}$

Now,

Suppose a line “s” perpendicular to given line l passing through point $P\left(1,2\right)$ .

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

Therefore,

Slope of line “s” is 1.

Equation of line “s”:

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

Now, on solving equation 1 and 2,

Point of intersection:

  $\begin{array}{l}\left(y=x+1\right)-\left(y=-x-2\right)\\ x=1.5\end{array}$

  $x=-1.5$

So,

  $\begin{array}{l}y=x+1\\ y=-1.5+1\\ y=-0.5\end{array}$

Point of intersection: $\left(-1.5,-0.5\right)$

Suppose the point of intersection is T $\left(-1.5,-0.5\right)$ .

Therefore,

The distance between the point $P\left(1,2\right)$ and T $\left(-1.5,-0.5\right)$ ,

Using distance formula:

  $\begin{array}{l}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ d=\sqrt{{\left(-1.5-1\right)}^2+{\left(-0.5-2\right)}^2}\\ d=\sqrt{12.5}\\ d=3.5\end{array}$

Thus, the required distance is 3.5 units.