Straight Lines – GUJCET Mathematics Study Notes
GUJCET (Gujarat Common Entrance Test) – Mathematics – Straight Lines
1. Simple Explanation
A straight line in the Cartesian plane is a set of points that satisfy a linear relation between the coordinates (x) and (y).
Mathematically it can be written in many equivalent forms:
| Form | Example | When to use |
|---|---|---|
| Slope–Intercept | (y = mx + c) | When slope (m) and (y)-intercept (c) are known. |
| Point–Slope | (y - y_1 = m(x - x_1)) | When one point ((x_1,y_1)) and slope (m) are known. |
| Two‑Point | (\displaystyle \frac{y-y_1}{y_2-y_1} = \frac{x-x_1}{x_2-x_1}) | When two distinct points ((x_1,y_1)) and ((x_2,y_2)) are known. |
| General | (ax + by + c = 0) | When coefficients are integers or fractions. |
| Intercept | (\displaystyle \frac{x}{a} + \frac{y}{b} = 1) | When (x)- and (y)-intercepts are known. |
| Normal (Distance) Form | (\displaystyle \frac{x\cos\theta + y\sin\theta}{p} = 1) | When distance (p) from origin and direction (\theta) are given. |
Key Idea – All these forms represent the same straight line; you can convert one to another using algebra.
2. Key Points
| Topic | Formula / Concept | Notes |
|---|---|---|
| Slope (m) | (m = \dfrac{y_2-y_1}{x_2-x_1}) | (m) is rise over run; vertical line → undefined. |
| y‑intercept (c) | Set (x=0) in (y=mx+c). | If line passes through ((x_0,y_0)): (c = y_0 - m x_0). |
| x‑intercept (a) | Set (y=0) → (x = -c/m) (if (m\neq 0)). | For intercept form (\frac{x}{a}+\frac{y}{b}=1). |
| Distance between parallel lines | If (ax+by+c_1=0) and (ax+by+c_2=0): | |
| (\displaystyle d = \frac{ | c_2-c_1 | }{\sqrt{a^2+b^2}}) |
| Distance from point ((x_0,y_0)) to line (ax+by+c=0) | (\displaystyle d = \frac{ | ax_0+by_0+c |
| Angle between two lines | For slopes (m_1,m_2): | |
| (\displaystyle \tan\theta = \left | \frac{m_2-m_1}{1+m_1m_2}\right | ) |
| Line perpendicular to given line | If line 1 has slope (m), line 2 has slope (-1/m) (unless (m=0) or undefined). | Use point‑slope or general form. |
| Line parallel to given line | Keep slope same, change intercept. | For general form: keep (a,b) same, change (c). |
| Normal form | (\displaystyle \frac{x\cos\theta + y\sin\theta}{p}=1) | (p) = perpendicular distance from origin; (\theta) = angle with positive (x)-axis. |
| Symmetry | Line (y=x) is the line of symmetry for a square or circle centered at origin. | Useful for reflections. |
3. Worked Example
Problem
Find the equation of the straight line:
- that passes through the points (P(2,5)) and (Q(6,1));
- find the distance from the point (R(4,0)) to this line;
- find the equation of the line perpendicular to this line and passing through (S(3,2));
- determine the distance between the original line and the perpendicular line.
Step‑by‑Step Solution
3.1 Equation of line (PQ)
-
Slope
[ m = \frac{y_Q-y_P}{x_Q-x_P} = \frac{1-5}{6-2} = \frac{-4}{4} = -1 ] -
Point‑Slope form (using (P(2,5)))
[ y-5 = -1(x-2) \