Class 10 › Mathematics › Chapter

Coordinate Geometry

Locating points algebraically in a plane: distance formula, section division, centroid, geometric shapes, and area of triangles.

1. Cartesian Plane, X and Y Axes

To locate the position of a point in a plane, we take two mutually perpendicular lines intersecting at their zero points:

  • X-axis: The horizontal number line $X'OX$.
  • Y-axis: The vertical number line $Y'OY$.
  • Origin ($O$): The intersection point $(0, 0)$.
  • Coordinates $(x, y)$: The abscissa ($x$) represents the perpendicular distance from the $y$-axis; the ordinate ($y$) represents the perpendicular distance from the $x$-axis.
X X' Y Y' O P(x, y) x (Abscissa) y (Ordinate)
Point $P(x, y)$ in the Cartesian plane with signed distances from axes.
2. Distance Formula with Proof

The distance between any two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ in the Cartesian plane is given by:

Distance Formula $$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
X Y O P(x₁, y₁) Q(x₂, y₂) M N R x₂ - x₁ y₂ - y₁ d
Geometric derivation of distance $PQ$ using right $\triangle PRQ$.
Proof
1. Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be two given points.
2. Draw perpendiculars $PM$ and $QN$ from $P$ and $Q$ to the $x$-axis.
3. From $P$, draw a horizontal line $PR$ perpendicular to $QN$ meeting $QN$ at $R$.
4. The coordinates of the feet on the $x$-axis are $OM = x_1$ and $ON = x_2$. Therefore: $$PR = MN = ON - OM = x_2 - x_1$$ 5. Similarly, along the vertical direction: $QN = y_2$ and $RN = PM = y_1$. Therefore: $$QR = QN - RN = y_2 - y_1$$ 6. Applying the **Pythagoras Theorem** in right-angled $\triangle PRQ$: $$PQ^2 = PR^2 + QR^2$$ $$PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$$ 7. Taking the positive square root (distance is always non-negative): $$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ Hence proved.
3. Distance of a Point from Origin

When finding the distance of any point $P(x, y)$ from the origin $O(0, 0)$, substitute $x_1 = 0, y_1 = 0, x_2 = x, y_2 = y$ into the distance formula:

Formula $$OP = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}$$
4. Properties of Common Quadrilaterals in Coordinate Geometry

To identify the type of quadrilateral formed by four given points $A, B, C, D$, calculate the lengths of the four sides ($AB, BC, CD, DA$) and the two diagonals ($AC, BD$):

(i) Parallelogram

Opposite sides are equal in length, but diagonals are unequal.

  • $AB = CD$ and $BC = DA$
  • Diagonals $AC \ne BD$
  • Alt check: Midpoint of $AC =$ Midpoint of $BD$

(ii) Rectangle

Opposite sides are equal and both diagonals are equal in length.

  • $AB = CD$ and $BC = DA$
  • Diagonals $AC = BD$
  • Midpoint of $AC =$ Midpoint of $BD$

(iii) Square

All four sides are equal and both diagonals are equal in length.

  • $AB = BC = CD = DA$
  • Diagonals $AC = BD$
  • Diagonals bisect at $90^\circ$

(iv) Rhombus

All four sides are equal, but diagonals are unequal in length.

  • $AB = BC = CD = DA$
  • Diagonals $AC \ne BD$
  • Midpoint of $AC =$ Midpoint of $BD$
5. Collinear Points (Using Distance Formula)

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

Condition using Distance Formula: Three points $A, B, C$ are collinear if and only if the sum of the lengths of any two segments equals the length of the third segment: $$AB + BC = AC \quad \text{or} \quad AB + AC = BC \quad \text{or} \quad AC + BC = AB$$
6. Section Formula (Internal Division)

The coordinates of point $P(x, y)$ which divides the line segment joining points $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m_1 : m_2$ are:

Section Formula $$P(x, y) = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \; \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right)$$
A(x₁, y₁) B(x₂, y₂) P(x, y) m₁ m₂
Point $P$ dividing segment $AB$ internally in ratio $AP : PB = m_1 : m_2$.
Ratio $k : 1$ Formulation: When determining an unknown ratio in which $P$ divides $AB$, assume the ratio to be $k : 1$. Then: $$P(x, y) = \left( \frac{k x_2 + x_1}{k + 1}, \; \frac{k y_2 + y_1}{k + 1} \right)$$
7. Midpoint Formula

The midpoint $M$ of a line segment divides it in the equal ratio $1 : 1$. Substituting $m_1 = 1$ and $m_2 = 1$ into the section formula gives:

Midpoint Formula $$M(x, y) = \left( \frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2} \right)$$
8. Centroid of a Triangle

The centroid ($G$) of a triangle is the point of concurrence of its three medians. It divides each median internally in the ratio $2 : 1$ (measuring from vertex to base).

G(x, y) A(x₁, y₁) B(x₂, y₂) C(x₃, y₃) D (Midpoint of BC)
Centroid $G$ dividing median $AD$ in ratio $AG : GD = 2 : 1$.
Derivation of Centroid Coordinates
1. Let $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$ be the vertices of $\triangle ABC$.
2. Median $AD$ bisects base $BC$. By the midpoint formula, the coordinates of $D$ are: $$D = \left( \frac{x_2 + x_3}{2}, \; \frac{y_2 + y_3}{2} \right)$$ 3. The centroid $G(x, y)$ divides $AD$ internally in the ratio $2 : 1$ ($m_1 = 2, m_2 = 1$).
4. Applying the section formula on $A(x_1, y_1)$ and $D$: $$x = \frac{2 \left(\frac{x_2 + x_3}{2}\right) + 1(x_1)}{2 + 1} = \frac{x_1 + x_2 + x_3}{3}$$ $$y = \frac{2 \left(\frac{y_2 + y_3}{2}\right) + 1(y_1)}{2 + 1} = \frac{y_1 + y_2 + y_3}{3}$$
Centroid Coordinates $$G(x, y) = \left( \frac{x_1 + x_2 + x_3}{3}, \; \frac{y_1 + y_2 + y_3}{3} \right)$$
9. Area of a Triangle with Proof

The area of $\triangle ABC$ formed by vertices $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ is:

Area Formula $$\text{Area}(\triangle ABC) = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$
X Y A(x₁, y₁) B(x₂, y₂) C(x₃, y₃) P Q R
Derivation by adding trapezia $ABPQ$ and $AQRC$, then subtracting trapezium $BPRC$.
Proof
1. Let $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$ be vertices of $\triangle ABC$.
2. Draw perpendiculars $BP, AQ, CR$ from vertices to the $x$-axis.
3. From the diagram, the area of $\triangle ABC$ is computed using three trapezia: $$\text{Area}(\triangle ABC) = \text{Area}(\text{trap } ABPQ) + \text{Area}(\text{trap } AQRC) - \text{Area}(\text{trap } BPRC)$$ 4. The area of a trapezium equals $\frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{distance between them})$: $$\text{Area}(\text{trap } ABPQ) = \frac{1}{2}(BP + AQ)(PQ) = \frac{1}{2}(y_2 + y_1)(x_1 - x_2)$$ $$\text{Area}(\text{trap } AQRC) = \frac{1}{2}(AQ + CR)(QR) = \frac{1}{2}(y_1 + y_3)(x_3 - x_1)$$ $$\text{Area}(\text{trap } BPRC) = \frac{1}{2}(BP + CR)(PR) = \frac{1}{2}(y_2 + y_3)(x_3 - x_2)$$ 5. Combining and expanding the terms: $$\text{Area}(\triangle ABC) = \frac{1}{2}\Big[(y_2 + y_1)(x_1 - x_2) + (y_1 + y_3)(x_3 - x_1) - (y_2 + y_3)(x_3 - x_2)\Big]$$ $$= \frac{1}{2}\Big[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\Big]$$ 6. Since area is a scalar measure and cannot be negative, we use the absolute value: $$\text{Area}(\triangle ABC) = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ Hence proved.
10. Finding Triangle Area Using the Ladder (Shoelace) Method

The Ladder Method (also known as the Shoelace Formula or Gauss's Area Formula) is an algorithmic, fast visual layout of the standard coordinate area formula. It arranges the vertex coordinates vertically in sequence and closes the loop by repeating the first vertex at the bottom.

Ladder Setup $$\text{Area}(\triangle ABC) = \frac{1}{2} \left| \; (x_1 y_2 + x_2 y_3 + x_3 y_1) - (y_1 x_2 + y_2 x_3 + y_3 x_1) \; \right|$$
x₁ y₁ x₂ y₂ x₃ y₃ x₁ y₁ + Down (↘) − Up (↗)
Shoelace cross-multiplication: Add downward products ($\searrow$) and subtract upward products ($\nearrow$).
Algorithm Steps
1. Write the coordinates in column pairs $(x, y)$ in counterclockwise or clockwise order: $(x_1, y_1), (x_2, y_2), (x_3, y_3)$.
2. Close the polygon: Repeat the initial vertex $(x_1, y_1)$ at the base of the stack.
3. Compute the downward diagonal sum ($S_{\downarrow}$): $$S_{\downarrow} = x_1 y_2 + x_2 y_3 + x_3 y_1$$ 4. Compute the upward diagonal sum ($S_{\uparrow}$): $$S_{\uparrow} = x_2 y_1 + x_3 y_2 + x_1 y_3$$ 5. Subtract $S_{\uparrow}$ from $S_{\downarrow}$, take the absolute value, and divide by $2$: $$\text{Area} = \frac{1}{2} \left| S_{\downarrow} - S_{\uparrow} \right|$$
Worked Example:

Find the area of $\triangle ABC$ whose vertices are $A(1, -1)$, $B(-4, 6)$, and $C(-3, -5)$.

Step 1: Set up the ladder $$\begin{matrix} x & y \\ \hline 1 & -1 \\ -4 & 6 \\ -3 & -5 \\ 1 & -1 \end{matrix}$$ Step 2: Calculate downward products ($S_{\downarrow}$) $$S_{\downarrow} = (1 \times 6) + ((-4) \times (-5)) + ((-3) \times (-1))$$ $$S_{\downarrow} = 6 + 20 + 3 = 29$$ Step 3: Calculate upward products ($S_{\uparrow}$) $$S_{\uparrow} = ((-1) \times (-4)) + (6 \times (-3)) + ((-5) \times 1)$$ $$S_{\uparrow} = 4 - 18 - 5 = -19$$ Step 4: Compute the area $$\text{Area} = \frac{1}{2} | S_{\downarrow} - S_{\uparrow} | = \frac{1}{2} | 29 - (-19) | = \frac{1}{2} | 29 + 19 | = \frac{1}{2} | 48 | = 24 \text{ sq. units}$$
10. Collinearity Condition Using Area of Triangle

If three points $A(x_1, y_1), B(x_2, y_2),$ and $C(x_3, y_3)$ lie on the same straight line, they cannot form a triangle. Thus, the area of the triangle formed by them must be zero.

Collinearity Condition $$x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0$$
Comparison of Collinearity Tests:
• Distance Method: Requires calculating 3 square roots ($AB, BC, AC$) and testing if the sum of two equals the third.
• Area Method: Requires purely polynomial arithmetic ($\text{Area} = 0$), avoiding radicals completely.