Chapter 13.1, Problem 13WE

To determine

To find AB, BC and AC

Answer to Problem 13WE

The values of $AB=2\sqrt{2},BC=5\sqrt{2}\text{ , }AC=3\sqrt{2}$ and ABC are collinear. A point lies between the other two points

Explanation of Solution

Given information:

The given points is

  $A\left(0,3\right)\text{ , }B\left(-2,1\right)\text{ , }C\left(3,6\right)$

Formula used:

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\left[d=\text{distance}\right]$

Calculation:

The distance between the two points can be calculated as

Let, $\left(x_1,y_1\right)\text{ be }\left(0,3\right)\text{ , }\left(x_2,y_2\right)\text{ be }\left(-2,1\right)\text{ and }\left(x_3,y_3\right)\text{ be }\left(3,6\right)$

Use the distance formula between A and B

  $AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\left[d=\text{distance}\right]$

Put the values of $x_1=0,x_2=-2,y_1=3\text{ and }{y}_2=1$ on the above equation

  $\begin{array}{l}AB=\sqrt{{\left(-2-0\right)}^2+{\left(1-3\right)}^2}\\ AB=\sqrt{{\left(-2\right)}^2+{\left(-2\right)}^2}\\ AB=\sqrt{4+4}\\ AB=\sqrt{8}\\ AB=2\sqrt{2}\end{array}$

Again, Use the distance formula between B and C

  $BC=\sqrt{{\left(x_3-x_2\right)}^2+{\left(y_3-y_2\right)}^2}\left[d=\text{distance}\right]$

Put the values of $x_2=-2,x_3=3,y_2=1\text{ and }{y}_3=6$ on the above equation

  $\begin{array}{l}BC=\sqrt{{\left(3-\left(-2\right)\right)}^2+{\left(6-1\right)}^2}\\ BC=\sqrt{{\left(3+2\right)}^2+{\left(6-1\right)}^2}\\ BC=\sqrt{{\left(5\right)}^2+{\left(5\right)}^2}\\ BC=\sqrt{25+25}\\ BC=\sqrt{50}\\ BC=5\sqrt{2}\end{array}$

Again, Use the distance formula between B and C

  $AC=\sqrt{{\left(x_3-x_1\right)}^2+{\left(y_3-y_1\right)}^2}\left[d=\text{distance}\right]$

Put the values of $x_1=0,x_3=3,y_1=3\text{ and }{y}_3=6$ on the above equation

  $\begin{array}{l}AC=\sqrt{{\left(3-0\right)}^2+{\left(6-3\right)}^2}\\ AC=\sqrt{{\left(3\right)}^2+{\left(3\right)}^2}\\ AC=\sqrt{9+9}\\ AC=\sqrt{18}\\ AC=3\sqrt{2}\end{array}$

Therefore,

  $\begin{array}{l}AB+AC=BC\\ BC=2\sqrt{2}+3\sqrt{2}\\ BC=5\sqrt{2}\end{array}$

Hence,

The values of $AB=2\sqrt{2},BC=5\sqrt{2}\text{ , }AC=3\sqrt{2}$ and ABC are collinear. A point lies between the other two points