Coordinate Geometry

Coordinate Geometry is a system where every point in the plane is described by an ordered pair (x, y). It allows us to analyze geometric shapes using algebra.

Think of it as placing geometry on graph paper, so that shapes and positions can be measured and calculated precisely.

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The Coordinate Plane

The plane is divided into 4 quadrants by:

• X-axis (horizontal): where $y = 0$
• Y-axis (vertical): where $x = 0$

Quadrants:

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Key Formulas and Concepts

1. Distance Between Two Points

Given $A(x_1, y_1)$ and $B(x_2, y_2)$:

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

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2. Section Formula

If point $P$ divides the line segment $AB$ in the ratio $m : n$, then:

$$P(x, y) = \left( \frac{mx_2 + nx_1}{m+n},\ \frac{my_2 + ny_1}{m+n} \right)$$

For internal division use $m:n$; for external division, adjust signs accordingly.

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3. Midpoint Formula

The midpoint $M$ of a line joining $A(x_1, y_1)$ and $B(x_2, y_2)$:

$$M = \left( \frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2} \right)$$

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4. Slope of a Line

Given two points $A(x_1, y_1)$, $B(x_2, y_2)$:

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

• Positive slope: line rises
• Negative slope: line falls
• Zero slope: horizontal line
• Undefined slope: vertical line

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5. Equation of a Line

Using slope and point $(x_1, y_1)$:

$$y - y_1 = m(x - x_1)$$

Two-point form:

$$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$$

General form:

$$Ax + By + C = 0$$

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6. Collinearity of Points

Three points $A, B, C$ are collinear if:

• The area of triangle ABC = 0
  or
• All three have the same slope between each pair

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7. Area of a Triangle (using coordinates)

Given three points: $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$

$$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$

If this area is 0, points are collinear.

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8. Locus

A locus is a set of all points satisfying a given condition. Example:

• Locus of all points equidistant from a fixed point = a circle
• Locus of all points equidistant from two fixed points = perpendicular bisector

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Visual Insights

Let’s say you have points:

• A(2, 3)
• B(6, 7)

To find the midpoint:

$$M = \left( \frac{2+6}{2}, \frac{3+7}{2} \right) = (4, 5)$$

To find slope:

$$m = \frac{7 - 3}{6 - 2} = \frac{4}{4} = 1$$

This line rises 1 unit for every 1 unit you move right — a 45° incline.

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Conceptual Tips and Common Mistakes

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Examples

Example 1

Find the distance between $A(1, 2)$ and $B(4, 6)$:

$$\sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

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Example 2

Find coordinates of the point dividing the line from $A(2, -1)$ to $B(4, 3)$ in the ratio 1:3.

$$x = \frac{(1)(4) + (3)(2)}{1 + 3} = \frac{4 + 6}{4} = 2.5 \quad y = \frac{(1)(3) + (3)(-1)}{4} = \frac{3 - 3}{4} = 0$$

Point = (2.5, 0)

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Example 3

Find the area of the triangle formed by points $A(0,0), B(4,0), C(4,3)$

$$\text{Area} = \frac{1}{2} \left| 0(0 - 3) + 4(3 - 0) + 4(0 - 0) \right| = \frac{1}{2} \times 12 = 6$$

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Example 4

Find the equation of a line passing through $(2, 5)$ with slope 3.

$$y - 5 = 3(x - 2) \Rightarrow y = 3x - 1$$

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Example 5

Are the points $A(1, 2), B(3, 6), C(5, 10)$ collinear?

Find slopes:

$$AB = \frac{6 - 2}{3 - 1} = 2, \quad BC = \frac{10 - 6}{5 - 3} = 2$$

Same slope ⇒ Collinear

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Advanced Insight: Reflection and Rotation

• Reflection across X-axis: $(x, y) \rightarrow (x, -y)$
• Reflection across Y-axis: $(x, y) \rightarrow (-x, y)$
• Rotation 90° counterclockwise about origin: $(x, y) \rightarrow (-y, x)$