# What is the centroid of a triangle with corners at (4 , 1 ), (3 , 2 ), and (5 , 0 )?

Jan 22, 2016

A triangle is formed by three non-collinear points.

But the given points are collinear therefore there is no triangle with these coordinates. And thus the question is meaningless,

If you have a question that how did I know that the given points are collinear then I am going to explain the answer.

Let $A \left({x}_{1} , {y}_{1}\right) , B \left({x}_{2} , {y}_{2}\right) \mathmr{and} C \left({x}_{3} , {y}_{3}\right)$ be three points then the condition for these three points to be collinear is that

$\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{{y}_{3} - {y}_{1}}{{x}_{3} - {x}_{1}}$

Here let $A = \left(4 , 1\right) , B = \left(3 , 2\right) \mathmr{and} C = \left(5 , 0\right)$

$\implies \frac{2 - 1}{3 - 4} = \frac{0 - 1}{5 - 4}$

$\implies \frac{1}{-} 1 = - \frac{1}{1}$

$\implies - 1 = - 1$

Since the condition is verified therefore the given points are collinear.

However if the man who gave you the question still says you to find the centroid then use the formula for finding the centroid which is used below.

If $A \left({x}_{,} {y}_{1}\right) , B \left({x}_{2} , {y}_{2}\right) \mathmr{and} C \left({x}_{3} , {y}_{3}\right)$ are the three vertices of a triangle it's centroid is given by

$G = \left(\frac{{x}_{1} + {x}_{2} + {x}_{3}}{3} , \frac{{y}_{1} + {y}_{2} + {y}_{3}}{3}\right)$

Where $G$ is the centroid

Here let $A = \left(4 , 1\right) , B = \left(3 , 2\right) \mathmr{and} C = \left(5 , 0\right)$

$\implies G = \left(\frac{4 + 3 + 5}{3} , \frac{1 + 2 + 0}{3}\right)$

$\implies G = \left(\frac{12}{3} , \frac{3}{3}\right)$

$\implies G = \left(4 , 1\right)$

Therefore, the centroid is $\left(4 , 1\right)$.