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?

16. A(3, 4), B(-3, 0), C(-1, 1)

Short answer

The values of$AB=2\sqrt{13},BC=\sqrt{5}\text{,}AC=5$ .

A, B, Care not collinear.

Step-by-step solution

Step-1 – Given

The given points are $A\left(3,4\right)\text{,}B\left(-3,0\right)\text{,}C\left(-1,1\right)$.

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:

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

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

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

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

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

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 $AB=2\sqrt{13},BC=\sqrt{5}\text{,}AC=5$.