Chapter 8

Coordinate Geometry

In-depth theory and application of linear equations and points.

1. Concept of Locus

A Locus is defined as the set of all points that satisfy a specific geometric condition. When a point moves under this condition, the path it traces is called its locus.

Examples:
• Locus of points at a distance 'r' from a fixed point: A Circle.
• Locus of points equidistant from two fixed points: The Perpendicular Bisector.

2. Section Formula

For a point \(P(x, y)\) dividing the line segment \(A(x_1, y_1)B(x_2, y_2)\) in ratio \(m_1 : m_2\):

A. Internal Division

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

B. External Division

When the point \(P\) lies outside the segment \(AB\), the formula uses subtraction:

\[x = \frac{m_1x_2 - m_2x_1}{m_1 - m_2}, \quad y = \frac{m_1y_2 - m_2y_1}{m_1 - m_2}\]

3 & 4. Slope and Slope-Intercept Form

The Slope (Gradient) \(m\) measures the inclination of a line.

\[m = \frac{y_2 - y_1}{x_2 - x_1} = \tan \theta\]

The Slope-Intercept Form:

\[y = mx + c\] Where \(m = \text{slope}\) and \(c = \text{y-intercept}\)

5 & 6. Double-Intercept and Normal Form

Double-Intercept Form

\[\frac{x}{a} + \frac{y}{b} = 1\] Where \(a\) = x-intercept, \(b\) = y-intercept

Normal (Perpendicular) Form

\[x \cos \alpha + y \sin \alpha = p\] Where \(p\) = perpendicular distance from origin, \(\alpha\) = angle of the normal with positive x-axis

7. Reduction of General Equation \(Ax + By + C = 0\)

A. Reduction to Slope-Intercept Form (\(y = mx + c\))

\[y = \left(-\frac{A}{B}\right)x + \left(-\frac{C}{B}\right)\] Slope \(m = -A/B\), y-intercept \(c = -C/B\)

B. Reduction to Point-Slope Form (\(y - y_1 = m(x - x_1)\))

\[y - y_1 = \left(-\frac{A}{B}\right)(x - x_1)\]

C. Reduction to Double-Intercept Form (\(\frac{x}{a} + \frac{y}{b} = 1\))

\[\frac{x}{(-C/A)} + \frac{y}{(-C/B)} = 1\] x-intercept \(a = -C/A\), y-intercept \(b = -C/B\)

D. Reduction to Normal (Perpendicular) Form (\(x \cos \alpha + y \sin \alpha = p\))

\[\left(\frac{A}{\sqrt{A^2+B^2}}\right)x + \left(\frac{B}{\sqrt{A^2+B^2}}\right)y = \frac{-C}{\sqrt{A^2+B^2}}\] Where \(\cos \alpha = \frac{A}{\sqrt{A^2+B^2}}\), \(\sin \alpha = \frac{B}{\sqrt{A^2+B^2}}\), and \(p = \frac{-C}{\sqrt{A^2+B^2}}\)