Question

Given points A, B, and C. Find AB, BC, and AC. Are A, B, and C collinear?

If so, which point lies between the other two?

15. A(-5, 6), B(0, 2), C(3, 0)

Short answer

The values of$AB=\sqrt{41},BC=\sqrt{13}\text{,}AC=10$ .

A, B, Care not collinear.

Step-by-step solution

Step-1 – Given

The given points are $A\left(-5,6\right)\text{,}B\left(0,2\right)\text{,}C\left(3,0\right)$.

Step-2 – To determine

We have to find the lengths AB, BCand AC.Then have to check ifA, B, C are collinear. And determine that which point lies between the other two.

Step-3 – Calculation 

Using the distance formula, the distance between A and B is:

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

Using the distance formula, the distance between B and C is:

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

Using the distance formula, the distance between A and C is:

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

Note that the sum of any two lines is not equal to length of other line.

So, A, B, Care not collinear.

The values of $AB=\sqrt{41},BC=\sqrt{13}\text{,}AC=10$.