Chapter 13.1, Problem 14WE

To determine

To find AB, BC and AC

Answer to Problem 14WE

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

Explanation of Solution

Given information:

The given points is

  $A\left(5,-5\right)\text{ , }B\left(0,5\right)\text{ , }C\left(2,1\right)$

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, $\left(x_1,y_1\right)\text{ be }\left(5,-5\right)\text{ , }\left(x_2,y_2\right)\text{ be }\left(0,5\right)\text{ and }\left(x_3,y_3\right)\text{ be }\left(2,1\right)$

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

  $\begin{array}{l}AB=\sqrt{{\left(0-5\right)}^2+{\left(5-\left(-5\right)\right)}^2}\\ AB=\sqrt{{\left(0-5\right)}^2+{\left(5+5\right)}^2}\\ AB=\sqrt{{\left(-5\right)}^2+{\left(10\right)}^2}\\ AB=\sqrt{25+100}\\ AB=\sqrt{125}\\ AB=5\sqrt{5}\end{array}$

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 $x_2=0,x_3=2,y_2=5\text{ and }{y}_3=1$ on the above equation

  $\begin{array}{l}BC=\sqrt{{\left(2-0\right)}^2+{\left(1-5\right)}^2}\\ BC=\sqrt{{\left(2\right)}^2+{\left(-4\right)}^2}\\ BC=\sqrt{4+16}\\ BC=\sqrt{20}\\ BC=2\sqrt{5}\end{array}$

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 $x_1=5,x_3=2,y_1=-5\text{ and }{y}_3=1$ on the above equation

  $\begin{array}{l}AC=\sqrt{{\left(2-5\right)}^2+{\left(1-\left(-5\right)\right)}^2}\\ AC=\sqrt{{\left(2-5\right)}^2+{\left(1+5\right)}^2}\\ AC=\sqrt{{\left(-3\right)}^2+{\left(6\right)}^2}\\ AC=\sqrt{9+36}\\ AC=\sqrt{45}\\ AC=3\sqrt{5}\end{array}$

Therefore,

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

Hence,

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