Question

The centroid of any triangle is located at the point of intersection of the medians.

Short answer

The centroid of any triangle is located at the point of intersection of the medians.

Step-by-step solution

Concept of median and midpoint formula

Median: A median is a line segment in a triangle from a vertex to the midpoint of the opposite side.

Midpoint formula: The midpoint can be found with the formula\(\frac{{\left( {{{(x)}_1} + {{(x)}_2}} \right)}}{2}\), \(\frac{{({{(y)}_1} + {{(y)}_2})}}{2}\). Here\({(x)_1},{(y)_1}\), and \({(x)_2},{(y)_2}\) are the coordinates of two points, and the midpoint is a point lying equidistant and between these two points.

Obtain midpoints by using the midpoint formula

Consider a triangle and place the coordinate axes in such a way that the vertices are \((a,0),(0,b)\), and \((c,0)\).

As the medians intersect at a point two-thirds of the way from each vertex to the opposite side, obtain the midpoints by the mid-point formula.

The midpoint between the points \((a,0)\) and \((0,b)\) is \(\left( {\frac{{a + 0}}{2},\frac{{0 + b}}{2}} \right) = \left( {\frac{a}{2},\frac{b}{2}} \right)\).

The midpoint between the points \((0,b)\) and \((c,0)\) is \(\left( {\frac{{0 + c}}{2},\frac{{b + 0}}{2}} \right) = \left( {\frac{c}{2},\frac{b}{2}} \right)\).

The midpoint between the points \((c,0)\) and \((a,0)\) is \(\left( {\frac{{c + a}}{2},\frac{{0 + 0}}{2}} \right) = \left( {\frac{{c + a}}{2},0} \right)\).

Obtain the point of intersection by using section formula.

Now, obtain the point of intersection between the points \((a,0)\) and \(\left( {\frac{c}{2},\frac{b}{2}} \right)\) with the help of section formula as the line is divided by other medians into ratio of 2: 1.

The section formula is given as \(\left( {\frac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_1} + {m_2}}},\frac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}}} \right)\)where\({m_1}:{m_2}\)is the ratio in which the line between the two points\(\left( {{x_1},{y_1}} \right)\)and\(\left( {{x_2},{y_2}} \right)\)is divided.

The point becomes\(\left( {\frac{{c + a}}{3},\frac{b}{3}} \right)\).

Also, obtain the point of intersection between the points \((0,b)\) and \(\left( {\frac{{c + a}}{2},0} \right)\) :

The point becomes\(\left( {\frac{{c + a}}{3},\frac{b}{3}} \right)\).

Obtain the point of intersection between the points \((c,0)\) and\(\left( {\frac{a}{2},\frac{b}{2}} \right)\) :

The point becomes\(\left( {\frac{{c + a}}{3},\frac{b}{3}} \right)\).

By all three efforts, the point of intersection is same that is \(\frac{2}{3}\) of the median line away from the vertices.

Therefore, the centroid of any triangle is located at the point of intersection of the medians.