Chapter 13.1, Problem 15WE

To determine

To find AB, BC and AC

Answer to Problem 15WE

The values of $AB=\sqrt{41},BC=\sqrt{13}\text{ , }AC=10$ and ABC are not collinear.

Explanation of Solution

Given information:

The given points is

  $A\left(-5,6\right)\text{ , }B\left(0,2\right)\text{ , }C\left(3,0\right)$

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, $\left(x_1,y_1\right)\text{ be }\left(-5,6\right)\text{ , }\left(x_2,y_2\right)\text{ be }\left(0,2\right)\text{ and }\left(x_3,y_3\right)\text{ be }\left(3,0\right)$

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 $x_1=-5,x_2=0,y_1=6\text{ and }{y}_2=2$ on the above equation

  $\begin{array}{l}AB=\sqrt{{\left(0-\left(-5\right)\right)}^2+{\left(2-6\right)}^2}\\ AB=\sqrt{{\left(0+5\right)}^2+{\left(2-6\right)}^2}\\ AB=\sqrt{{\left(5\right)}^2+{\left(-4\right)}^2}\\ AB=\sqrt{25+16}\\ AB=\sqrt{41}\end{array}$

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 $x_2=0,x_3=3,y_2=2\text{ and }{y}_3=0$ on the above equation

  $\begin{array}{l}BC=\sqrt{{\left(3-0\right)}^2+{\left(0-2\right)}^2}\\ BC=\sqrt{{\left(3\right)}^2+{\left(-2\right)}^2}\\ BC=\sqrt{9+4}\\ BC=\sqrt{13}\end{array}$

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 $x_1=-5,x_3=3,y_1=6\text{ and }{y}_3=0$ on the above equation

  $\begin{array}{l}AC=\sqrt{{\left(3-\left(-5\right)\right)}^2+{\left(0-6\right)}^2}\\ AC=\sqrt{{\left(3+5\right)}^2+{\left(0-6\right)}^2}\\ AC=\sqrt{{\left(8\right)}^2+{\left(-6\right)}^2}\\ AC=\sqrt{64+36}\\ AC=\sqrt{100}\\ AC=10\end{array}$

Therefore,

Sum of any two lines not equal to length of other line

Hence,

The values of $AB=\sqrt{41},BC=\sqrt{13}\text{ , }AC=10$ and ABC are not collinear.