Coordinate Geometry: Collinearity, Trisection, and Proofs Class 10

In 1637, French philosopher and mathematician René Descartes lay in bed watching a fly crawl across a tiled ceiling. He realized that the insect's exact location could be described by two simple numbers: its perpendicular distance from two intersecting grid lines. That insight gave birth to Cartesian Coordinate Geometry—the mathematical bridge that transforms abstract geometric shapes into precise algebraic numbers.

In CBSE Class 10 Mathematics, Chapter 7 (Coordinate Geometry), questions in Section C and Section D demand more than basic formula substitution. You are asked to prove that three points are collinear, find the coordinates of points of trisection, and prove whether a given set of four vertices forms a rectangle, rhombus, or square.

In this guide, we master these advanced geometric proofs and section formula applications.

What You Will Learn

• Three methods to test Collinearity of Three Points (Distance, Section, and Area)
• How to divide a line segment into three equal parts (Points of Trisection)
• Determining the ratio in which coordinate axes divide a line segment
• Proving types of quadrilaterals algebraically: Parallelogram vs. Rectangle vs. Rhombus vs. Square
• The Mid-Point Diagonal Property for finding missing vertices
• Solved CBSE board examination problems and arithmetic shortcuts

1. Testing Collinearity of Three Points

Three points $A, B,$ and $C$ are said to be collinear if they all lie on the exact same straight line.

                         A -------------- B -------------- C
                         <------ AB -----><------ BC ----->
                         <----------------- AC ------------>

Method 1: The Distance Method

Three points $A, B,$ and $C$ are collinear if the sum of the lengths of any two segments equals the length of the third segment: $\mathbf{AB + BC = AC}$

• Calculate $AB, BC,$ and $AC$ using the Distance Formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
• If the two smaller distances add up to the largest distance, the points are collinear!

Method 2: The Section Ratio Method

Assume that point $B(x_2, y_2)$ divides the segment joining $A(x_1, y_1)$ and $C(x_3, y_3)$ in the ratio $k : 1$.

• Solve for $k$ using the $x$-coordinate equation: $x_2 = \frac{k x_3 + x_1}{k + 1}$.
• Substitute this calculated value of $k$ into the $y$-coordinate equation. If it yields the true value of $y_2$, the points are collinear!

2. Points of Trisection of a Line Segment

To trisect a line segment means to divide it into three equal parts:

                         A -------- P -------- Q -------- B
                         <-- 1 part -><-- 1 part -><-- 1 part ->
                         P divides AB in ratio 1 : 2
                         Q divides AB in ratio 2 : 1 (or Q is mid-point of PB!)

The Trisection Algorithm:

Let $P$ and $Q$ be the two points of trisection of line segment $AB$:

1. Point $P$ divides segment $AB$ internally in the ratio $1 : 2$. Apply the section formula with $m_1 = 1, m_2 = 2$: $\mathbf{P = \left( \frac{1(x_2) + 2(x_1)}{1 + 2}, \; \frac{1(y_2) + 2(y_1)}{1 + 2} \right)}$
2. Point $Q$ divides segment $AB$ internally in the ratio $2 : 1$ (or simply use the midpoint formula on $P$ and $B$!): $\mathbf{Q = \left( \frac{x_P + x_B}{2}, \; \frac{y_P + y_B}{2} \right)}$

Solved Example: Finding Points of Trisection (NCERT Classic)

Problem: Find the coordinates of the points of trisection of the line segment joining $A(2, -2)$ and $B(-7, 4)$.

Solution:

1. Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be the points of trisection.
2. Find Point $P$ (divides $AB$ in ratio $1 : 2$): $x_1 = \frac{1(-7) + 2(2)}{1 + 2} = \frac{-7 + 4}{3} = \frac{-3}{3} = \mathbf{-1}$ $y_1 = \frac{1(4) + 2(-2)}{1 + 2} = \frac{4 - 4}{3} = \frac{0}{3} = \mathbf{0}$ Therefore, $P = (-1, 0)$.
3. Find Point $Q$ (Mid-point of $PB$): $x_2 = \frac{x_P + x_B}{2} = \frac{-1 + (-7)}{2} = \frac{-8}{2} = \mathbf{-4}$ $y_2 = \frac{y_P + y_B}{2} = \frac{0 + 4}{2} = \frac{4}{2} = \mathbf{2}$ Therefore, $Q = (-4, 2)$.
4. <u>The points of trisection are $(-1, 0)$ and $(-4, 2)$</u>.

3. Proving Quadrilateral Types via Coordinate Geometry

In 4-mark questions, you are given four points and asked to determine the exact geometric quadrilateral they form:

                            Quadrilateral Decision Tree
                                         |
       +---------------------------------+---------------------------------+
       |                                                                   |
ALL 4 SIDES EQUAL (AB = BC = CD = DA)                       OPPOSITE SIDES EQUAL (AB = CD, BC = DA)
       |                                                                   |
   +---+---+                                                           +---+---+
   |       |                                                           |       |
Diagonals  Diagonals                                                Diagonals  Diagonals
EQUAL      NOT Equal                                                EQUAL      NOT Equal
   |       |                                                           |       |
SQUARE     RHOMBUS                                                  RECTANGLE  PARALLELOGRAM

The Examiner's Golden Rule: <u>To prove a square or rectangle, proving side lengths is NOT enough! You MUST also calculate the lengths of both diagonals ($AC$ and $BD$)! If diagonals are equal, it is a square/rectangle; if diagonals are unequal, it is a rhombus/parallelogram!</u>

4. Finding Missing Vertices Using Diagonal Bisectors

A classic 3-mark problem gives three vertices of a parallelogram $A, B, C$ and asks for the fourth vertex $D(x, y)$:

• The Core Geometric Property: The diagonals of a parallelogram bisect each other, meaning the mid-point of diagonal $AC$ is IDENTICAL to the mid-point of diagonal $BD$! $\mathbf{\left( \frac{x_A + x_C}{2}, \; \frac{y_A + y_C}{2} \right) = \left( \frac{x_B + x_D}{2}, \; \frac{y_B + y_D}{2} \right)}$ Equating the respective $x$ and $y$ coordinates solves for $D(x, y)$ in seconds, completely avoiding messy distance formula quadratics!

5. Summary and Examination Tips

Exam Tip: When asked for the ratio in which the $x$-axis divides a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$, use the shortcut ratio $-y_1 : y_2$! For the $y$-axis, the ratio is $-x_1 : x_2$!

Common Mistake: In the section formula, multiplying $m_1$ with $x_1$. Remember: $m_1$ cross-multiplies with $x_2$, and $m_2$ cross-multiplies with $x_1$!