Equation of a Line

ICSE Class 10 Mathematics — Chapter 14 Master Notes

1. Basic Concept & Linear Equations

A linear equation in two variables of first degree ($ax + by + c = 0$) represents a straight line in a Cartesian coordinate plane.

Fundamental Axiom of Line Equations:
1. Any point whose coordinates satisfy the equation of a line lies on that line.
2. Any point through which a line passes will always satisfy the equation of that line.

Master Method: Testing if Point Lies on Line / Finding Parameter:

To check if $P(x_1, y_1)$ lies on line $ax + by + c = 0$, substitute $x = x_1$ and $y = y_1$. If $\text{LHS} = \text{RHS}$, the point lies on the line. If a point $(k, 2)$ lies on $3x - 8y = 2$, substitute $x=k, y=2 \implies 3k - 16 = 2 \implies k = 6$.

2. Inclination & Slope (Gradient) of a Line

1. Inclination of a Line ($\theta$): The angle $\theta$ which the line makes with the positive direction of the $x$-axis, measured in the anti-clockwise direction ($0^\circ \le \theta < 180^\circ$).

Inclination of $x$-axis (and line parallel to $x$-axis) $= 0^\circ$.
Inclination of $y$-axis (and line parallel to $y$-axis) $= 90^\circ$.

2. Slope or Gradient ($m$): The tangent of inclination $\theta$ of a line is called its slope or gradient, denoted by $m$.

$$m = \tan \theta$$

Slope of $x$-axis: $m = \tan 0^\circ = 0$.
Slope of $y$-axis: $m = \tan 90^\circ = \text{undefined}$ ($\infty$).
If $\theta$ is acute ($0^\circ < \theta < 90^\circ$), slope $m$ is **positive**.
If $\theta$ is obtuse ($90^\circ < \theta < 180^\circ$), slope $m$ is **negative**.

3. Slope of a Line Passing Through Two Points

Slope Formula Through Two Points:

The slope $m$ of a line passing through points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is:

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{y_1 - y_2}{x_1 - x_2} = \frac{\text{Difference of Ordinates}}{\text{Difference of Abscissae}}$$

4. Parallel and Perpendicular Lines

Slope Conditions for Parallel and Perpendicular Lines:

Parallel Lines ($L_1 \parallel L_2$): Two non-vertical lines are parallel if and only if their slopes are equal.
$$m_1 = m_2$$
Perpendicular Lines ($L_1 \perp L_2$): Two non-vertical lines are perpendicular if and only if the product of their slopes is $-1$ (negative reciprocal).
$$m_1 \times m_2 = -1 \implies m_2 = -\frac{1}{m_1}$$

5. Collinearity of Three Points

Three or more points $A, B$, and $C$ are said to be collinear if they lie on the same straight line.

Condition: Points $A, B$, and $C$ are collinear if and only if:

$$\text{Slope of } AB = \text{Slope of } BC = \text{Slope of } AC$$

6. Various Forms of Equations of Straight Lines

Standard Forms of Straight Line Equations:

Slope-Intercept Form: Line with slope $m$ and $y$-intercept $c$:
$$y = mx + c$$
Point-Slope Form: Line passing through $(x_1, y_1)$ with slope $m$:
$$y - y_1 = m(x - x_1)$$
Two-Point Form: Line passing through $(x_1, y_1)$ and $(x_2, y_2)$:
$$y - y_1 = \left(\frac{y_2 - y_1}{x_2 - x_1}\right)(x - x_1)$$
Intercept Form: Line making $x$-intercept $a$ and $y$-intercept $b$:
$$\frac{x}{a} + \frac{y}{b} = 1$$
Equally Inclined Lines: Lines making equal angles with coordinate axes ($\theta = 45^\circ \implies m=1$ or $\theta = 135^\circ \implies m=-1$):
$$y = x + c \quad \text{or} \quad y = -x + c$$

7. Extracting Slope and Intercepts from General Equation $ax + by + c = 0$

Converting general form $ax + by + c = 0$ into slope-intercept form $y = mx + c$:

$$by = -ax - c \implies y = \left(-\frac{a}{b}\right)x + \left(-\frac{c}{b}\right)$$

Slope ($m$): $$m = -\frac{a}{b} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } y}$$
Y-Intercept ($c$): $$c = -\frac{c}{b} = -\frac{\text{Constant term}}{\text{Coefficient of } y}$$
X-Intercept ($a$): Set $y = 0 \implies x = -\frac{c}{a} = -\frac{\text{Constant term}}{\text{Coefficient of } x}$$

8. Step-by-Step Solved Master Examples & Interactive Chart

Interactive Line Plot: Parallel & Perpendicular Lines

Visualizing Given Line $L_1: 2x + 3y + 4 = 0$ (Slope $-2/3$) and Perpendicular Line $L_2: 3x - 2y = 12$ (Slope $3/2$).

Example 1 (Finding Slope & Inclination): Find the slope and inclination of the line passing through points $A(-2, 3)$ and $B(2, 7)$.

Solution:

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 3}{2 - (-2)} = \frac{4}{4} = 1$$

Since $m = \tan \theta = 1 \implies \theta = 45^\circ$.

Answer: Slope $= 1$, Inclination $= 45^\circ$.

Example 2 (Equation of Parallel Line): Find the equation of a line passing through $(2, -1)$ and parallel to the line $2x - y = 4$.

Solution:

Convert given line $2x - y = 4 \implies y = 2x - 4$. Slope $m = 2$.

Since the required line is parallel, its slope $m = 2$.

Using Point-Slope form with $(x_1, y_1) = (2, -1)$:

$$y - (-1) = 2(x - 2) \implies y + 1 = 2x - 4 \implies y = 2x - 5 \quad \text{or} \quad 2x - y - 5 = 0$$

Example 3 (Equation of Perpendicular Line): Find the equation of the line passing through $(-2, 3)$ and perpendicular to the line $2x + 3y + 4 = 0$.

Solution:

Slope of given line $2x + 3y + 4 = 0$ is $m_1 = -\frac{a}{b} = -\frac{2}{3}$.

For perpendicular line, $m_2 = -\frac{1}{m_1} = \frac{3}{2}$.

Using Point-Slope form with $(x_1, y_1) = (-2, 3)$:

$$y - 3 = \frac{3}{2}(x - (-2)) \implies 2(y - 3) = 3(x + 2)$$ $$2y - 6 = 3x + 6 \implies 3x - 2y + 12 = 0 \quad \text{or} \quad 2y = 3x + 12$$

Example 4 (Perpendicular Bisector of Segment): Given $A(-5, 2)$ and $B(1, -4)$. Find the equation of the perpendicular bisector of segment $AB$.

Solution:

1. Midpoint $P$ of $AB$:

$$P = \left( \frac{-5 + 1}{2}, \frac{2 + (-4)}{2} \right) = \left( \frac{-4}{2}, \frac{-2}{2} \right) = (-2, -1)$$

2. Slope of segment $AB$:

$$m_{AB} = \frac{-4 - 2}{1 - (-5)} = \frac{-6}{6} = -1$$

3. Slope of perpendicular bisector $m = -\frac{1}{-1} = 1$.

4. Equation passing through $P(-2, -1)$ with slope $m = 1$:

$$y - (-1) = 1(x - (-2)) \implies y + 1 = x + 2 \implies y = x + 1$$

Example 5 (Rhombus Diagonal Equation): $ABCD$ is a rhombus. Vertices $A$ and $C$ are $(3, 6)$ and $(-1, 2)$ respectively. Find the equation of diagonal $BD$.

Solution:

Diagonals of a rhombus bisect each other at right angles ($90^\circ$).

1. Intersection point $P$ is the midpoint of $AC$:

$$P = \left( \frac{3 + (-1)}{2}, \frac{6 + 2}{2} \right) = (1, 4)$$

2. Slope of diagonal $AC$:

$$m_{AC} = \frac{2 - 6}{-1 - 3} = \frac{-4}{-4} = 1$$

3. Slope of diagonal $BD$ (perpendicular to $AC$): $m_{BD} = -1$.

4. Equation of $BD$ through $P(1, 4)$ with slope $-1$:

$$y - 4 = -1(x - 1) \implies y - 4 = -x + 1 \implies x + y = 5$$

9. Quick Revision Formula Matrix

Target Concept	Given Input	Direct Formula
Slope ($m$)	Inclination $\theta$	$m = \tan \theta$
Slope ($m$)	Two points $(x_1,y_1), (x_2,y_2)$	$m = \frac{y_2-y_1}{x_2-x_1}$
Parallel Condition	Slopes $m_1, m_2$	$m_1 = m_2$
Perpendicular Condition	Slopes $m_1, m_2$	$m_1 \times m_2 = -1$
Slope from General Form	Equation $ax + by + c = 0$	$m = -\frac{a}{b}$
Slope-Intercept Form	Slope $m$, y-intercept $c$	$y = mx + c$
Point-Slope Form	Slope $m$, Point $(x_1,y_1)$	$y - y_1 = m(x - x_1)$

10. ICSE Board PYQs & High-Yield Practice Problems

ICSE Board PYQ 2015

BOARD Three vertices of parallelogram $ABCD$ taken in order are $A(3, 6)$, $B(5, 10)$, and $C(3, 2)$. Find:

(i) Coordinates of fourth vertex $D$.
(ii) Length of diagonal $BD$.
(iii) Equation of side $AB$.

ICSE Board PYQ 2013

BOARD In right $\triangle ABC$ at $B$, $AB = 6\text{ cm}, DE = 4\text{ cm}, AC = 15\text{ cm}$ with $DE \perp BC$. Find the ratio of areas $\text{Area}(\triangle ABC) : \text{Area}(\triangle DEC)$.

ICSE Board PYQ 2012

BOARD Find the equation of the line passing through $(-2, 1)$ and perpendicular to the line $4x + 5y = 6$.

ICSE Board PYQ 2011

BOARD $ABCD$ is a parallelogram where $A(x, y)$, $B(5, 8)$, $C(4, 7)$, and $D(2, -4)$. Find coordinates of $A$ and equation of diagonal $BD$.

ICSE Board PYQ 2010

BOARD The equation of a line is $3x + 4y - 7 = 0$. Find:

(i) The slope of the line.
(ii) The equation of a line perpendicular to it passing through intersection of $x - y + 2 = 0$ and $3x + y - 10 = 0$.

ICSE Board PYQ 2008

BOARD $A(7, -3)$ and $B(1, 9)$ are given points. Find:

(i) Slope of $AB$.
(ii) Equation of perpendicular bisector of $AB$.
(iii) Value of $p$ if $(-2, p)$ lies on the bisector.

ICSE Board PYQ 2005

BOARD Line $AB$ passing through $A(0, 3)$ and $B(3, 4)$ intersects $CD$ at $P(3, 4)$. Find equations of $AB$ and $CD$ if inclination of $CD$ is $60^\circ$.