Chapter 13.1, Problem 16WE

To determine

To find AB, BC and AC

Answer to Problem 16WE

The values of $AB=2\sqrt{13},BC=\sqrt{5}\text{ , }AC=5$ and ABC are not collinear.

Explanation of Solution

Given information:

The given points is

  $A\left(3,4\right)\text{ , }B\left(-3,0\right)\text{ , }C\left(-1,1\right)$

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, $\left(x_1,y_1\right)\text{ be }\left(3,4\right)\text{ , }\left(x_2,y_2\right)\text{ be }\left(-3,0\right)\text{ and }\left(x_3,y_3\right)\text{ be }\left(-1,1\right)$

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 $x_1=3,x_2=-3,y_1=4\text{ and }{y}_2=0$ on the above equation

  $\begin{array}{l}AB=\sqrt{{\left(-3-3\right)}^2+{\left(0-4\right)}^2}\\ AB=\sqrt{{\left(-6\right)}^2+{\left(-4\right)}^2}\\ AB=\sqrt{36+16}\\ AB=\sqrt{52}\\ AB=2\sqrt{13}\end{array}$

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 $x_2=-3,x_3=-1,y_2=0\text{ and }{y}_3=1$ on the above equation

  $\begin{array}{l}BC=\sqrt{{\left(-1-\left(-3\right)\right)}^2+{\left(1-0\right)}^2}\\ BC=\sqrt{{\left(-1+3\right)}^2+{\left(1-0\right)}^2}\\ BC=\sqrt{{\left(2\right)}^2+{\left(1\right)}^2}\\ BC=\sqrt{4+1}\\ BC=\sqrt{5}\end{array}$

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 $x_1=3,x_3=-1,y_1=4\text{ and }{y}_3=1$ on the above equation

  $\begin{array}{l}AC=\sqrt{{\left(-1-3\right)}^2+{\left(1-4\right)}^2}\\ AC=\sqrt{{\left(-4\right)}^2+{\left(-3\right)}^2}\\ AC=\sqrt{16+9}\\ AC=\sqrt{25}\\ AC=5\end{array}$

Therefore,

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

Hence,

The values of $AB=2\sqrt{13},BC=\sqrt{5}\text{ , }AC=5$ and ABC are not collinear.