The Mid-Point Formula and Centroid of a Triangle for CBSE Class 10

Finding the exact center of a line segment or the balance point of a geometric triangle is one of the most common requirements in physics, engineering, and coordinate geometry. While the general section formula handles arbitrary division ratios $m_1 : m_2$, the most frequent and elegant case occurs when a segment is bisected into two exactly equal halves ($1 : 1$ ratio).

In CBSE Class 10 Mathematics, Chapter 7 (Coordinate Geometry) uses the Mid-Point Formula to solve high-frequency board exam problems involving circles (center as midpoint of diameter), parallelograms (diagonals bisecting each other), and the Centroid of a Triangle.

What You Will Learn

• Derivation of the Mid-Point Formula as a special case of the section formula
• Using the Mid-Point Formula to find the center and radius of a circle from its diameter
• The Parallelogram Diagonals Property (finding missing vertex coordinates without distance formula)
• Definition of a median and the Centroid of a Triangle
• Formula for the Centroid: $G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$
• Solved CBSE board examination problems and shortcut strategies

1. The Mid-Point Formula

Derivation:

Consider a line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$. Let $M(x, y)$ be the mid-point of $AB$. Since $M$ bisects $AB$, it divides the segment in the equal ratio $1 : 1$ ($m_1 = 1, m_2 = 1$).

Substituting $m_1 = 1$ and $m_2 = 1$ into the section formula: $x = \frac{1(x_2) + 1(x_1)}{1 + 1} = \frac{x_1 + x_2}{2}$ $y = \frac{1(y_2) + 1(y_1)}{1 + 1} = \frac{y_1 + y_2}{2}$

The Mid-Point Formula

The coordinates of the mid-point $M$ of the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ are: $M(x, y) = \left( \frac{x_1 + x_2}{2}, \quad \frac{y_1 + y_2}{2} \right)$

$\text{Mid-point} = \left( \text{Average of } x\text{-coordinates}, \quad \text{Average of } y\text{-coordinates} \right)$

2. High-Yield Application 1: Circle Diameter Problems

In a circle, the diameter passes through the center, and the center is the exact mid-point of the diameter.

Solved Example: Finding Coordinates of a Point on a Circle

Problem: Find the coordinates of a point $A$, where $AB$ is the diameter of a circle whose centre is $(2, -3)$ and $B$ is $(1, 4)$.

Solution:

1. Let the coordinates of point $A$ be $(x, y)$.
2. The centre $C(2, -3)$ is the mid-point of diameter $AB$, where $B = (1, 4)$.
3. Apply the mid-point formula: $\frac{x + 1}{2} = 2 \quad \text{and} \quad \frac{y + 4}{2} = -3$
4. Solve each equation: $x + 1 = 4 \implies x = 3$ $y + 4 = -6 \implies y = -10$
5. Therefore, <u>the coordinates of point $A$ are $(3, -10)$</u>.

3. High-Yield Application 2: Parallelogram Diagonals Shortcut

A fundamental theorem of Euclidean geometry states:

The diagonals of a parallelogram bisect each other.

This means that for any parallelogram $ABCD$, the mid-point of diagonal $AC$ is identical to the mid-point of diagonal $BD$!

                         A -------------------- B
                          \                    /
                           \        O         /
                            \   (Mid-point)  /
                             \              /
                              D ---------- C
                   Mid-point of AC = Mid-point of BD

Important: <u>Whenever a board problem gives three vertices of a parallelogram and asks for the missing fourth vertex, NEVER use the long distance formula! Always equate the midpoints of the two diagonals. It solves the entire problem in under two minutes!</u>

Solved Example: Finding Unknown Vertex and Parameter

Problem: If the points $A(6, 1)$, $B(8, 2)$, $C(9, 4)$, and $D(p, 3)$ are the vertices of a parallelogram, taken in order, find the value of $p$.

Solution:

1. Since $ABCD$ is a parallelogram, its diagonals $AC$ and $BD$ bisect each other at the same point $O$.
2. Mid-point of diagonal $AC$: $O = \left( \frac{6 + 9}{2}, \; \frac{1 + 4}{2} \right) = \left( \frac{15}{2}, \; \frac{5}{2} \right)$
3. Mid-point of diagonal $BD$: $O = \left( \frac{8 + p}{2}, \; \frac{2 + 3}{2} \right) = \left( \frac{8 + p}{2}, \; \frac{5}{2} \right)$
4. Since both midpoints represent the exact same point, equate their $x$-coordinates: $\frac{8 + p}{2} = \frac{15}{2}$ $8 + p = 15 \implies p = 15 - 8 = 7$
5. Therefore, <u>the value of $p$ is $7$</u>.

4. The Centroid of a Triangle

A median of a triangle is a line segment joining a vertex to the mid-point of the opposite side. Every triangle has three medians.

Definition of Centroid

The point of concurrence where all three medians of a triangle intersect is called the Centroid of the triangle, denoted by $G$. The centroid represents the center of gravity of the triangle and divides each median internally in the ratio $2 : 1$ (from vertex to base).

                                      A(x1, y1)
                                     /|                                     / |                                     /  |                                     /   G (2:1)
                                 /    |                                     B-----D------C
                              (D is mid-point of BC)

Derivation of the Centroid Formula:

Let the vertices of $\Delta ABC$ be $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$.

1. Mid-point $D$ of base $BC$: $D = \left( \frac{x_2 + x_3}{2}, \; \frac{y_2 + y_3}{2} \right)$
2. Centroid $G$ divides median $AD$ internally in the ratio $2 : 1$ ($m_1 = 2, m_2 = 1$).
3. Apply the Section Formula: $x_G = \frac{2\left(\frac{x_2 + x_3}{2}\right) + 1(x_1)}{2 + 1} = \frac{(x_2 + x_3) + x_1}{3} = \frac{x_1 + x_2 + x_3}{3}$ $y_G = \frac{2\left(\frac{y_2 + y_3}{2}\right) + 1(y_1)}{2 + 1} = \frac{(y_2 + y_3) + y_1}{3} = \frac{y_1 + y_2 + y_3}{3}$

The Centroid Formula

The coordinates of the centroid $G$ of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ are: $G(x, y) = \left( \frac{x_1 + x_2 + x_3}{3}, \quad \frac{y_1 + y_2 + y_3}{3} \right)$

5. Solved Example: Centroid Calculation

Problem: Find the coordinates of the centroid of a triangle whose vertices are $A(-3, 0)$, $B(5, -2)$, and $C(-8, 5)$.

Solution:

1. Apply the centroid formula: $x_G = \frac{x_1 + x_2 + x_3}{3} = \frac{-3 + 5 + (-8)}{3} = \frac{-6}{3} = -2$ $y_G = \frac{y_1 + y_2 + y_3}{3} = \frac{0 + (-2) + 5}{3} = \frac{3}{3} = 1$
2. Therefore, <u>the coordinates of the centroid are $(-2, 1)$</u>.

6. Summary and Examination Tips

Exam Tip: In parallelogram vertex questions, clearly state: "Since the diagonals of a parallelogram bisect each other, the midpoint of AC coincides with the midpoint of BD."

Common Mistake: In the centroid formula, dividing by $2$ instead of $3$. A midpoint averages $2$ points (divide by $2$); a centroid averages $3$ vertices (divide by $3$!).