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📚 MATHSKILLER TEXTBOOK SERIES
05

Coordinate Geometry

© 2025 MathsKiller
CHAPTER 5

Coordinate Geometry

🎯 Learning Objectives

5.1
Basic Formulas
NameFormula
Distance$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
Midpoint$M = \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$
Slope$m = \dfrac{y_2-y_1}{x_2-x_1}$
5.2
Equations of Straight Lines
FormFormulaKey Feature
Slope-intercept$y = mx + c$slope = $m$, y-int = $c$
Point-slope$y - y_1 = m(x - x_1)$passes $(x_1, y_1)$
General$ax + by + c = 0$slope = $-\frac{a}{b}$

⚡ Parallel & Perpendicular Conditions

Parallel: $m_1 = m_2$

Perpendicular: $m_1 \times m_2 = -1$

5.3
Point-to-Line Distance
$d = \dfrac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$
5.4
Equations of Circles
FormFormula
Standard$(x-h)^2 + (y-k)^2 = r^2$
General$x^2 + y^2 + Dx + Ey + F = 0$

Centre: $(-\frac{D}{2}, -\frac{E}{2})$, Radius: $r = \sqrt{\frac{D^2}{4} + \frac{E^2}{4} - F}$

📋 DSE Past Paper Questions

DSE 2020 Q16

The distance from $(3, 4)$ to the line $3x + 4y - 10 = 0$ is?

A. 1   B. 3   C. 5   D. 15

📝 Solution

$d = \dfrac{|3(3) + 4(4) - 10|}{\sqrt{9 + 16}} = \dfrac{15}{5} = 3$

Answer: B

📝 Practice Questions

1. Find the distance between A(1, 3) and B(7, 11).
2. Find the equation of the line through (3, -1) with slope 2.
3. Find the centre and radius of $x^2 + y^2 + 2x - 8y - 8 = 0$.

📋 Answers

1. 10   2. $y = 2x - 7$   3. Centre $(-1, 4)$, $r = 5$

📘 MathsKiller Textbook Series | Chapter 5: Coordinate Geometry

© 2025 MathsKiller