Coordinate Geometry

Class 10 Mathematics • Chapter 07 (Rationalized Syllabus)

1. Introduction to Cartesian Plane

Coordinate System: Two perpendicular number lines intersecting at origin (0,0) divide the plane into 4 quadrants.

x-coordinate (Abscissa): Distance from y-axis.
y-coordinate (Ordinate): Distance from x-axis.
Point on x-axis: $(x, 0)$
Point on y-axis: $(0, y)$

2. Distance Formula

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

DISTANCE FORMULA

$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Note: Distance is always non-negative.

AI IMAGE PROMPT:
Generate a professional educational diagram on a pure white background in landscape.
Subject: Cartesian Plane with X and Y axes. Plot two points P(x₁, y₁) and Q(x₂, y₂) in the first quadrant.
Details: Draw a straight line connecting P and Q. Draw a right-angled triangle PQR where R is (x₂, y₁) to visualize the derivation of the distance formula using Pythagoras Theorem.
Style: Clean black lines, Serif font for labels.

Special Applications

Distance from Origin: Distance of $P(x, y)$ from origin $(0,0)$ is $\sqrt{x^2 + y^2}$.
Equidistant Points: If P is equidistant from A and B, then $PA = PB$ or $PA^2 = PB^2$. (Squaring removes the root).
Collinear Points: Three points A, B, and C are collinear if $AB + BC = AC$ (Also area of triangle = 0, but Area method is deleted).

3. Section Formula

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

SECTION FORMULA

$$ x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \quad y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} $$

AI IMAGE PROMPT:
Generate a professional geometry diagram on a pure white background in landscape.
Subject: A line segment with endpoints A(x₁, y₁) and B(x₂, y₂). A point P(x, y) lies on the segment, dividing it into two parts.
Labels: Label the length AP as corresponding to ratio m₁ and PB as ratio m₂.
Style: Minimalist, high contrast, educational style.

Mid-Point Formula

If P is the mid-point of AB, then ratio is $1:1$.

$$ Mid\text{-}Point = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) $$

4. Important Exam Problems (PYQ Trends)

Type 1: Finding an Equidistant Point

Q: Find a point on the x-axis which is equidistant from $(2, -5)$ and $(-2, 9)$.

Method:

Let point on x-axis be $P(x, 0)$. Let $A(2, -5)$ and $B(-2, 9)$.
Given $PA = PB \Rightarrow PA^2 = PB^2$.
$(x-2)^2 + (0-(-5))^2 = (x-(-2))^2 + (0-9)^2$.
$(x-2)^2 + 25 = (x+2)^2 + 81$.
$x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$.
$-4x + 29 = 4x + 85$.
$-8x = 56 \Rightarrow x = -7$.
Point is $(-7, 0)$.

Type 2: The Parallelogram Problem

Q: If $(1, 2), (4, y), (x, 6)$ and $(3, 5)$ are vertices of a parallelogram taken in order, find x and y.

Logic: Diagonals of a parallelogram bisect each other. So, Mid-point of AC = Mid-point of BD.

AI IMAGE PROMPT:
Generate a diagram of a Parallelogram labeled ABCD in counter-clockwise order.
Details: Draw diagonals AC and BD intersecting at point O. Mark that AO=OC and BO=OD (indicating bisection).
Background: Pure white. Style: Thin black lines.

Mid-point of AC = $(\frac{1+x}{2}, \frac{2+6}{2}) = (\frac{1+x}{2}, 4)$
Mid-point of BD = $(\frac{4+3}{2}, \frac{y+5}{2}) = (\frac{7}{2}, \frac{y+5}{2})$
Equating them: $\frac{1+x}{2} = \frac{7}{2} \Rightarrow 1+x = 7 \Rightarrow x = 6$.
$4 = \frac{y+5}{2} \Rightarrow 8 = y+5 \Rightarrow y = 3$.

EXAM TRICK: FINDING RATIO k:1 When ratio is unknown, NEVER assume it as $m_1:m_2$ (two variables). Always assume it as $k:1$ (one variable). Use one coordinate to find k, then find the other coordinate.

SYLLABUS NOTE Area of Triangle formula ($\frac{1}{2}[x_1(y_2-y_3)...]$) is DELETED from the rationalized syllabus. Collinearity must be checked using the Distance Formula ($AB+BC=AC$).