Q. 14 - Geometry

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?

14. A(5, -5), B(0, 5), C(2, 1)

Step-by-Step Solution

Answer

The values of$AB=5\sqrt{5},BC=2\sqrt{5}\text{ , }AC=3\sqrt{5}$ .

 A, B, C are collinear. 

Point C lies between the other two points.

Step-1 – Given

The given points are $A\left(5,-5\right)\text{ , }B\left(0,5\right)\text{ , }C\left(2,1\right)$.

Step-2 – To determine

We have to find the lengths AB, BC and AC. Then have to check if A, 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:

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

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

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

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

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

Hence,

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

So, the values of $AB=5\sqrt{5},BC=2\sqrt{5}\text{ , }AC=3\sqrt{5}$.

 A, B, C are collinear. 

Point C lies between the other two points.