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WBB • Class XI • Mathematics • Ch 11
Estimated Time: 80 minutes
Study Progress: In Progress

Straight Lines

The straight line is the simplest, most foundational geometric locus in coordinate geometry, embodying constant direction and uniform rate of change across the Cartesian plane. First formalized through René Descartes analytical geometry in 1637, coordinate geometry establishes an exact algebraic representation of geometric curves. A straight line is formally defined as the locus of a point moving such that the gradient or slope between any two arbitrary points on it remains invariant. In this chapter, students explore the geometric intuition of inclination, the trigonometric definition of slope as m = tan θ, and the exact criteria for parallelism and perpendicularity. The curriculum advances through diverse canonical representations: point-slope form, two-point form, slope-intercept form, intercept form, normal form, and the symmetric parametric distance form. Furthermore, the general linear equation Ax + By + C = 0 is systematically decomposed into these canonical models. The chapter provides comprehensive analytical mechanisms for calculating the acute and obtuse angles between intersecting lines, perpendicular distances from points to lines, distances between parallel lines, the concurrency of three coplanar lines, the locus of a family of lines passing through a fixed intersection, and the exact coordinates of the foot of a perpendicular and specular reflection image across a linear mirror.

Why This Chapter Matters

Straight lines form the bedrock of applied linear algebra, computer graphics, optimization, robotics, optical ray tracing, and physical modeling. In computer graphics and gaming engines, rendering algorithms—such as Bresenham line algorithm and ray-casting—rely directly on the parametric and slope-intercept equations of straight lines to project three-dimensional scenes onto two-dimensional pixel arrays. In robotics and automated navigation, path planning and obstacle detection algorithms represent motion trajectories and boundary walls as line segments and calculate shortest distances using the perpendicular distance formula. In economics and operations research, linear programming optimizes production allocations by determining feasible regions bounded by straight-line inequalities. In optics, the law of reflection and refraction is analyzed by finding the image of light rays and calculating reflections across linear interfaces. For students preparing for WBCHSE Class 11 annual examinations and competitive engineering entrances including WBJEE, JEE Main, and JEE Advanced, Straight Lines is a foundational, high-scoring chapter that directly bridges coordinate geometry, vectors, and differential calculus.

Chapter Roadmap & Progression

1 1. Fundamentals: Cartesian Plane, I...
2 2. Canonical Forms of the Equation...
3 3. Normal (Perpendicular) Form & Pa...
4 4. General Equation of a Line & Can...
5 5. Angle Between Two Lines, Paralle...
6 6. Distance Formulas, Family of Lin...

Complete Concept Guide (100% Curriculum Coverage)

1. Fundamentals: Cartesian Plane, Inclination, Slope & Collinearity

Coordinate geometry unites algebra and Euclidean geometry by expressing geometric points as ordered real number pairs $(x, y)$ and loci as algebraic equations.

1.1 Recall of 2D Coordinate Geometry
  • Distance Formula: The distance between $P(x_1, y_1)$ and $Q(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
  • Section Formula: The point $R$ dividing the line segment joining $P(x_1, y_1)$ and $Q(x_2, y_2)$ in the ratio $m : n$ is: $$\text{Internal: } \left(\frac{m x_2 + n x_1}{m + n}, \; \frac{m y_2 + n y_1}{m + n}\right), \quad \text{External: } \left(\frac{m x_2 - n x_1}{m - n}, \; \frac{m y_2 - n y_1}{m - n}\right)$$
  • Area of Triangle & Collinearity: Area of $\triangle ABC$ with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is: $$\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$ Points $A, B, C$ are collinear if and only if $\Delta = 0$.
1.2 Inclination and Slope (Gradient) of a Line
Inclination ($\theta$): The angle $\theta$ which a straight line makes with the positive direction of the $x$-axis measured in the anticlockwise direction is called its inclination. The range of inclination is: $$0^\circ \le \theta < 180^\circ \quad (\text{or } 0 \le \theta < \pi)$$

Slope ($m$): If $\theta$ is the inclination of a non-vertical straight line ($\theta \neq 90^\circ$), then the trigonometrical tangent of this angle is defined as the slope or gradient of the line:

$$\mathbf{m = \tan \theta}$$
  • For a line parallel to the $x$-axis (horizontal line), $\theta = 0^\circ \implies m = \tan 0^\circ = \mathbf{0}$.
  • For a line parallel to the $y$-axis (vertical line), $\theta = 90^\circ \implies m = \tan 90^\circ$ is undefined (or infinite).
  • If $\theta$ is acute ($0^\circ < \theta < 90^\circ$), the slope $m > 0$ (positive slope).
  • If $\theta$ is obtuse ($90^\circ < \theta < 180^\circ$), the slope $m < 0$ (negative slope).
1.3 Slope of a Line Passing Through Two Given Points

Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ with $x_1 \neq x_2$ be two distinct points on a non-vertical line of inclination $\theta$. Constructing a right-angled triangle with horizontal base $(x_2 - x_1)$ and vertical rise $(y_2 - y_1)$:

$$\mathbf{m = \tan \theta = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{Change in } y}{\text{Change in } x}}$$

Collinearity via Slopes: Three points $A, B, C$ are collinear if and only if $\mathbf{\text{Slope}(AB) = \text{Slope}(BC)}$.

2. Canonical Forms of the Equation of a Straight Line

Depending on the geometric parameters specified (points, slopes, intercepts), the equation of a straight line can be expressed in various standard forms.

2.1 Lines Parallel to the Coordinate Axes
  • Line parallel to the $x$-axis: Every point on the line has a constant $y$-coordinate. Equation: $\mathbf{y = k}$ (for $x$-axis itself, $\mathbf{y = 0}$).
  • Line parallel to the $y$-axis: Every point on the line has a constant $x$-coordinate. Equation: $\mathbf{x = c}$ (for $y$-axis itself, $\mathbf{x = 0}$).
2.2 Point-Slope Form

The equation of a straight line passing through a fixed point $(x_1, y_1)$ and having slope $m$ is derived from the slope definition for any arbitrary point $(x, y)$ on the line:

$$\frac{y - y_1}{x - x_1} = m \implies \mathbf{y - y_1 = m(x - x_1)}$$
2.3 Two-Point Form

The equation of a straight line passing through two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ is obtained by substituting $m = \frac{y_2 - y_1}{x_2 - x_1}$ into the point-slope form:

$$\mathbf{y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1)}$$
2.4 Slope-Intercept Form

Let a line have slope $m$ and intersect the $y$-axis at $(0, c)$, so that its $y$-intercept is $c$. Substituting $(0, c)$ into point-slope form gives:

$$\mathbf{y = mx + c}$$

If the line makes an $x$-intercept $d$, intersecting the $x$-axis at $(d, 0)$, its equation is $\mathbf{y = m(x - d)}$.

2.5 Intercept Form

Let a line make non-zero intercepts $a$ on the $x$-axis and $b$ on the $y$-axis, passing through $(a, 0)$ and $(0, b)$. Using two-point form:

$$y - 0 = \frac{b - 0}{0 - a} (x - a) \implies y = -\frac{b}{a}(x - a) \implies \frac{y}{b} = -\frac{x}{a} + 1$$
$$\mathbf{\frac{x}{a} + \frac{y}{b} = 1}$$ Note: $a$ is the directed length from origin to $x$-intercept; $b$ is the directed length to $y$-intercept.

3. Normal (Perpendicular) Form & Parametric Distance Form

When the position of a line is defined through origin-directed perpendicular rays or along-line distances, the normal and parametric forms provide unmatched analytical power.

3.1 Normal (Perpendicular) Form

Let $p$ be the length of the perpendicular (normal) drawn from the origin $O(0, 0)$ to a straight line $L$, and let $\alpha$ be the angle which this normal makes with the positive direction of the $x$-axis ($0 \le \alpha < 360^\circ$ and $p \ge 0$).

The foot of the perpendicular $N$ on the line has polar-to-Cartesian coordinates:

$$N = (p \cos \alpha, \; p \sin \alpha)$$

Since the line $L$ is perpendicular to $ON$, its slope is $m_L = -\frac{1}{m_{ON}} = -\frac{1}{\tan \alpha} = -\frac{\cos \alpha}{\sin \alpha}$.

Using the point-slope form with point $N(p\cos\alpha, p\sin\alpha)$:

$$y - p \sin \alpha = -\frac{\cos \alpha}{\sin \alpha} (x - p \cos \alpha)$$ $$y \sin \alpha - p \sin^2 \alpha = -x \cos \alpha + p \cos^2 \alpha$$ $$x \cos \alpha + y \sin \alpha = p (\cos^2 \alpha + \sin^2 \alpha)$$
$$\mathbf{x \cos \alpha + y \sin \alpha = p}$$ where $p \ge 0$ is the non-negative perpendicular distance from the origin and $0 \le \alpha < 2\pi$.
3.2 Parametric (Symmetric / Distance) Form

Let a line pass through a given point $P(x_1, y_1)$ and make an inclination $\theta$ with the positive $x$-axis. If $Q(x, y)$ is any point on the line at an algebraic distance $r$ from $P$:

$$\cos \theta = \frac{x - x_1}{r}, \quad \sin \theta = \frac{y - y_1}{r}$$
$$\mathbf{\frac{x - x_1}{\cos \theta} = \frac{y - y_1}{\sin \theta} = r}$$ The coordinates of any point $Q$ at a distance $r$ from $P(x_1, y_1)$ along the line are: $$\mathbf{x = x_1 + r \cos \theta, \quad y = y_1 + r \sin \theta}$$ Distance Sign Convention: $r$ is positive for points in one direction along the line and negative for points in the opposite direction.

4. General Equation of a Line & Canonical Transformations

Every first-degree equation in two variables represents a straight line, and conversely, every straight line can be represented by a linear equation.

4.1 The General Linear Equation
Theorem: The general first-degree equation in $x$ and $y$: $$\mathbf{Ax + By + C = 0}$$ where $A, B, C \in \mathbb{R}$ and $A, B$ are not simultaneously zero ($A^2 + B^2 \neq 0$), always represents a straight line in the Cartesian plane.
4.2 Transformation into Standard Forms
Standard Form Algebraic Transformation of $Ax + By + C = 0$ Key Parameters
Slope-Intercept Form $$y = \left(-\frac{A}{B}\right)x + \left(-\frac{C}{B}\right) \quad (B \neq 0)$$ $\text{Slope } m = -\frac{A}{B}$
$\text{Intercept } c = -\frac{C}{B}$
Intercept Form $$\frac{x}{-C/A} + \frac{y}{-C/B} = 1 \quad (A, B, C \neq 0)$$ $x\text{-intercept } a = -\frac{C}{A}$
$y\text{-intercept } b = -\frac{C}{B}$
Normal Form $$\frac{Ax}{\pm\sqrt{A^2 + B^2}} + \frac{By}{\pm\sqrt{A^2 + B^2}} = \frac{-C}{\pm\sqrt{A^2 + B^2}}$$ Sign rule: Choose $\pm$ opposite to the sign of $C$ so RHS is $\ge 0$. $p = \frac{|C|}{\sqrt{A^2 + B^2}} \ge 0$
$\cos \alpha = \frac{-A}{\pm\sqrt{A^2+B^2}}$
$\sin \alpha = \frac{-B}{\pm\sqrt{A^2+B^2}}$

5. Angle Between Two Lines, Parallelism & Perpendicularity

When two lines intersect, they form two pairs of vertically opposite angles (one acute, one obtuse), completely determined by their slopes.

5.1 Angle Between Two Intersecting Lines

Let $L_1$ and $L_2$ be two lines with slopes $m_1$ and $m_2$ and inclinations $\theta_1$ and $\theta_2$ respectively ($\theta_2 > \theta_1$). The angle of intersection is $\theta = \theta_2 - \theta_1$.

$$\tan \theta = \tan(\theta_2 - \theta_1) = \frac{\tan \theta_2 - \tan \theta_1}{1 + \tan \theta_1 \tan \theta_2} = \frac{m_2 - m_1}{1 + m_1 m_2}$$
Acute Angle Formula: $$\mathbf{\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|} \quad (1 + m_1 m_2 \neq 0)$$ The obtuse angle between the lines is $\phi = 180^\circ - \theta$.
5.2 Conditions of Parallelism and Perpendicularity
  • Condition of Parallelism ($\theta = 0^\circ$): $$\tan 0^\circ = 0 \implies m_2 - m_1 = 0 \implies \mathbf{m_1 = m_2}$$ For general equations $A_1 x + B_1 y + C_1 = 0$ and $A_2 x + B_2 y + C_2 = 0$: $$\frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2}$$ Equation of any line parallel to $Ax + By + C = 0$ is $\mathbf{Ax + By + k = 0}$.
  • Condition of Perpendicularity ($\theta = 90^\circ$): $$\cot 90^\circ = 0 \implies 1 + m_1 m_2 = 0 \implies \mathbf{m_1 m_2 = -1} \quad \left(\text{or } m_2 = -\frac{1}{m_1}\right)$$ For general equations $A_1 x + B_1 y + C_1 = 0$ and $A_2 x + B_2 y + C_2 = 0$: $$A_1 A_2 + B_1 B_2 = 0$$ Equation of any line perpendicular to $Ax + By + C = 0$ is $\mathbf{Bx - Ay + k = 0}$.

6. Distance Formulas, Family of Lines, Concurrency & Reflections

The final module addresses spatial metric relations: perpendicular distances, bundles of concurrent lines, and geometric reflection transformations.

6.1 Perpendicular Distance of a Point from a Line
Theorem: The perpendicular length $d$ from a given point $P(x_1, y_1)$ to the line $Ax + By + C = 0$ is: $$\mathbf{d = \frac{|A x_1 + B y_1 + C|}{\sqrt{A^2 + B^2}}}$$ Distance from Origin $(0, 0)$: $\mathbf{d_0 = \frac{|C|}{\sqrt{A^2 + B^2}}}$.
6.2 Distance Between Two Parallel Lines

If two parallel lines are written with identical coefficients of $x$ and $y$ as $L_1: Ax + By + C_1 = 0$ and $L_2: Ax + By + C_2 = 0$:

$$\mathbf{d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}}$$ Crucial Warning: Always ensure that the coefficients of $x$ and $y$ in both lines are made identical before subtracting $C_1$ and $C_2$!
6.3 Family of Lines Passing Through the Intersection of Two Lines

Let $L_1 \equiv A_1 x + B_1 y + C_1 = 0$ and $L_2 \equiv A_2 x + B_2 y + C_2 = 0$ be two intersecting lines. The equation of any line passing through their point of intersection is:

$$\mathbf{L_1 + \lambda L_2 = 0 \iff (A_1 x + B_1 y + C_1) + \lambda (A_2 x + B_2 y + C_2) = 0}$$

where $\lambda \in \mathbb{R}$ is an arbitrary parameter determined by an additional geometric constraint.

6.4 Concurrency of Three Straight Lines

Three lines $A_i x + B_i y + C_i = 0$ ($i = 1, 2, 3$) are concurrent (pass through a common single point) if and only if:

$$\mathbf{\begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{vmatrix} = 0}$$
6.5 Foot of Perpendicular and Image (Reflection) of a Point
Let $P(x_1, y_1)$ be a point and $L: ax + by + c = 0$ be a line mirror:
  • Foot of Perpendicular $(h, k)$: $$\mathbf{\frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{a x_1 + b y_1 + c}{a^2 + b^2}}$$
  • Image / Reflection $(x', y')$: $$\mathbf{\frac{x' - x_1}{a} = \frac{y' - y_1}{b} = -2\frac{a x_1 + b y_1 + c}{a^2 + b^2}}$$

Key Formulas, Identities & Theorems

Two-Point Slope Formula
m = (y₂ - y₁) / (x₂ - x₁)
Inclination 0° ≤ θ < 180°. Parallel lines: m₁ = m₂; Perpendicular: m₁·m₂ = -1.
Slope-Intercept & Intercept Forms
y = mx + c and x/a + y/b = 1
c is y-intercept; a and b are non-zero intercepts on x-axis and y-axis.
Normal (Perpendicular) Form
x cos α + y sin α = p
p is perpendicular distance from origin; α is the normal inclination angle.
Angle Between Two Lines
tan θ = |(m₂ - m₁) / (1 + m₁m₂)|
Obtuse angle is 180° - θ. If 1 + m₁m₂ = 0, lines are mutually perpendicular (θ = 90°).
Perpendicular Distance from Point to Line
|Ax₁ + By₁ + C| / √(A² + B²)
Distance from origin is |C| / √(A² + B²).
Distance Between Parallel Lines
|C₁ - C₂| / √(A² + B²)
Coefficients of x and y must be identical in both equations before applying formula.

Conceptual Solved Examples & Case Studies

Example 1
Find the equation of a straight line passing through the point \((3, -2)\) and having slope \(-\frac{4}{3}\). [2 marks]
Step-by-Step Solution:
Solution: Given: Point $(x_1, y_1) = (3, -2)$ Slope $m = -\frac{4}{3}$ Using the Point-Slope Form of a line: $$y - y_1 = m(x - x_1)$$ Substitute $(x_1, y_1) = (3, -2)$ and $m = -\frac{4}{3}$: $$y - (-2) = -\frac{4}{3}(x - 3)$$ $$y + 2 = -\frac{4}{3}(x - 3)$$ Multiply both sides by $3$: $$3(y + 2) = -4(x - 3)$$ $$3y + 6 = -4x + 12$$ Rearranging into general form $Ax + By + C = 0$: $$4x + 3y + 6 - 12 = 0 \implies \mathbf{4x + 3y - 6 = 0}$$ Hence, the required equation of the straight line is $4x + 3y - 6 = 0$.
Example 2
Find the equation of a line whose perpendicular distance from the origin is \(4\) units and the angle which the normal makes with positive \(x\)-axis is \(15^\circ\). [3 marks]
Step-by-Step Solution:
Solution: Given: Perpendicular distance from origin $p = 4$ Normal inclination angle $\alpha = 15^\circ$ Using the Normal Form of a straight line: $$x \cos \alpha + y \sin \alpha = p$$ $$x \cos 15^\circ + y \sin 15^\circ = 4$$
Evaluate $\cos 15^\circ$ and $\sin 15^\circ$: $$\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$$ $$\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$$
Substitute back into normal form: $$x \left(\frac{\sqrt{3} + 1}{2\sqrt{2}}\right) + y \left(\frac{\sqrt{3} - 1}{2\sqrt{2}}\right) = 4$$ Multiply both sides by $2\sqrt{2}$: $$(\sqrt{3} + 1)x + (\sqrt{3} - 1)y = 8\sqrt{2}$$ Hence, the equation of the line is $\mathbf{(\sqrt{3} + 1)x + (\sqrt{3} - 1)y - 8\sqrt{2} = 0}$.
Example 3
Find the perpendicular distance between the parallel lines \(3x - 4y + 9 = 0\) and \(6x - 8y - 15 = 0\). [3 marks]
Step-by-Step Solution:
Solution: Given lines: $$L_1: 3x - 4y + 9 = 0 \quad \text{--- (Equation 1)}$$ $$L_2: 6x - 8y - 15 = 0 \quad \text{--- (Equation 2)}$$ To apply the parallel distance formula, the coefficients of $x$ and $y$ in both equations must be identical. Multiply Equation 1 by $2$: $$2(3x - 4y + 9) = 0 \implies 6x - 8y + 18 = 0$$ Now comparing: Line 1: $6x - 8y + 18 = 0 \implies A = 6, \; B = -8, \; C_1 = 18$ Line 2: $6x - 8y - 15 = 0 \implies A = 6, \; B = -8, \; C_2 = -15$ Using the distance formula between parallel lines: $$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$$ Substitute the values: $$d = \frac{|18 - (-15)|}{\sqrt{6^2 + (-8)^2}} = \frac{|18 + 15|}{\sqrt{36 + 64}} = \frac{33}{\sqrt{100}} = \frac{33}{10} = \mathbf{3.3 \text{ units}}$$ Hence, the distance between the parallel lines is $3.3$ units (or $\frac{33}{10}$ units).
Example 4
Find the coordinates of the foot of the perpendicular and the image of the point \((1, 2)\) with respect to the line \(x - 3y + 4 = 0\). [4 marks]
Step-by-Step Solution:
Solution: Given line: $x - 3y + 4 = 0 \implies a = 1, \; b = -3, \; c = 4$ Given point: $(x_1, y_1) = (1, 2)$ Evaluate $ax_1 + by_1 + c$ and $a^2 + b^2$: $$ax_1 + by_1 + c = 1(1) - 3(2) + 4 = 1 - 6 + 4 = -1$$ $$a^2 + b^2 = 1^2 + (-3)^2 = 1 + 9 = 10$$
Part 1: Foot of Perpendicular $(h, k)$ Using the standard formula: $$\frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{a x_1 + b y_1 + c}{a^2 + b^2}$$ $$\frac{h - 1}{1} = \frac{k - 2}{-3} = -\frac{-1}{10} = \frac{1}{10}$$ Solving for $h$: $$h - 1 = \frac{1}{10} \implies h = 1 + \frac{1}{10} = \mathbf{\frac{11}{10}}$$ Solving for $k$: $$k - 2 = -3 \times \frac{1}{10} = -\frac{3}{10} \implies k = 2 - \frac{3}{10} = \mathbf{\frac{17}{10}}$$ Hence, the foot of the perpendicular is $\mathbf{\left(\frac{11}{10}, \frac{17}{10}\right)}$.
Part 2: Image / Reflection $(x', y')$ Using the standard reflection formula: $$\frac{x' - x_1}{a} = \frac{y' - y_1}{b} = -2 \frac{a x_1 + b y_1 + c}{a^2 + b^2}$$ $$\frac{x' - 1}{1} = \frac{y' - 2}{-3} = -2\left(-\frac{1}{10}\right) = \frac{2}{10} = \frac{1}{5}$$ Solving for $x'$: $$x' - 1 = \frac{1}{5} \implies x' = 1 + \frac{1}{5} = \mathbf{\frac{6}{5}}$$ Solving for $y'$: $$y' - 2 = -3 \times \frac{1}{5} = -\frac{3}{5} \implies y' = 2 - \frac{3}{5} = \mathbf{\frac{7}{5}}$$ Hence, the image of the point $(1, 2)$ in the line is $\mathbf{\left(\frac{6}{5}, \frac{7}{5}\right)}$.
Example 5
Find the equation of the straight line passing through the intersection of the lines \(2x + 3y - 4 = 0\) and \(x - 5y + 7 = 0\) and perpendicular to the line \(3x + 4y - 5 = 0\). [4 marks]
Step-by-Step Solution:
Solution:
Step 1: Family of lines through intersection The equation of any line passing through the intersection of $2x + 3y - 4 = 0$ and $x - 5y + 7 = 0$ is: $$(2x + 3y - 4) + \lambda(x - 5y + 7) = 0$$ Regrouping terms into $Ax + By + C = 0$: $$(2 + \lambda)x + (3 - 5\lambda)y + (-4 + 7\lambda) = 0 \quad \text{--- (Equation 1)}$$ The slope of this line is: $$m_1 = -\frac{\text{Coefficient of } x}{\text{Coefficient of } y} = -\frac{2 + \lambda}{3 - 5\lambda} = \frac{2 + \lambda}{5\lambda - 3}$$
Step 2: Perpendicularity condition The given line is $3x + 4y - 5 = 0$. Its slope is: $$m_2 = -\frac{3}{4}$$ Since the required line is perpendicular to $3x + 4y - 5 = 0$: $$m_1 \cdot m_2 = -1$$ $$\left(\frac{2 + \lambda}{5\lambda - 3}\right) \left(-\frac{3}{4}\right) = -1$$ Multiply both sides by $-1$: $$\frac{3(2 + \lambda)}{4(5\lambda - 3)} = 1$$ $$3(2 + \lambda) = 4(5\lambda - 3)$$ $$6 + 3\lambda = 20\lambda - 12$$ $$6 + 12 = 20\lambda - 3\lambda$$ $$18 = 17\lambda \implies \mathbf{\lambda = \frac{18}{17}}$$
Step 3: Finding the line equation Substitute $\lambda = \frac{18}{17}$ into Equation 1: $$\left(2 + \frac{18}{17}\right)x + \left(3 - 5\cdot\frac{18}{17}\right)y + \left(-4 + 7\cdot\frac{18}{17}\right) = 0$$ $$\left(\frac{34 + 18}{17}\right)x + \left(\frac{51 - 90}{17}\right)y + \left(\frac{-68 + 126}{17}\right) = 0$$ $$\frac{52}{17}x - \frac{39}{17}y + \frac{58}{17} = 0$$ Multiply by $17$: $$\mathbf{52x - 39y + 58 = 0}$$ Notice that $52/13 = 4$ and $39/13 = 3$, so the slope is $52/39 = 4/3$, which satisfies $m_1 \cdot (-3/4) = -1$. Hence, the required equation is $52x - 39y + 58 = 0$.
Example 6
(i) A ray of light passing through the point \((1, 2)\) reflects on the \(x\)-axis at point \(A\) and the reflected ray passes through the point \((5, 3)\). Find the coordinates of \(A\). [2.5 marks] (ii) Find the value of \(k\) for which the three lines \(3x + y - 2 = 0\), \(kx + 2y - 3 = 0\), and \(2x - y - 3 = 0\) are concurrent. [2.5 marks] [5 marks total]
Step-by-Step Solution:
Solution:
Part (i): Ray of light reflection on the $x$-axis: [2.5 marks] Let the incident ray pass through $P(1, 2)$ and strike the $x$-axis at point $A(x_0, 0)$. The reflected ray passes through $Q(5, 3)$. By the optical law of reflection, the image of point $P(1, 2)$ in the reflecting line ($x$-axis, $y = 0$) will lie on the line containing the reflected ray $AQ$. The image of $P(1, 2)$ across the $x$-axis is: $$P'(1, -2)$$ Since $P'$, $A$, and $Q$ are collinear, the slope of $P'A$ equals the slope of $P'Q$: $$\text{Slope}(P'Q) = \frac{3 - (-2)}{5 - 1} = \frac{3 + 2}{4} = \frac{5}{4}$$ Now equation of line $P'Q$: $$y - (-2) = \frac{5}{4}(x - 1) \implies y + 2 = \frac{5}{4}(x - 1)$$ $$4(y + 2) = 5(x - 1) \implies 4y + 8 = 5x - 5 \implies 5x - 4y - 13 = 0$$ Since point $A(x_0, 0)$ lies on the $x$-axis, set $y = 0$: $$5x_0 - 4(0) - 13 = 0 \implies 5x_0 = 13 \implies \mathbf{x_0 = \frac{13}{5}}$$ Hence, the coordinates of point $A$ are $\mathbf{\left(\frac{13}{5}, 0\right)}$.
Part (ii): Value of $k$ for concurrency of three lines: [2.5 marks] Given lines: $$L_1: 3x + y - 2 = 0$$ $$L_2: kx + 2y - 3 = 0$$ $$L_3: 2x - y - 3 = 0$$ Three lines are concurrent if and only if the determinant formed by their coefficients vanishes: $$\begin{vmatrix} 3 & 1 & -2 \\ k & 2 & -3 \\ 2 & -1 & -3 \end{vmatrix} = 0$$ Expanding along the first row: $$3 [2(-3) - (-3)(-1)] - 1 [k(-3) - (-3)(2)] + (-2) [k(-1) - 2(2)] = 0$$ $$3 [-6 - 3] - 1 [-3k + 6] - 2 [-k - 4] = 0$$ $$3(-9) + 3k - 6 + 2k + 8 = 0$$ $$-27 + 5k + 2 = 0$$ $$5k - 25 = 0 \implies 5k = 25 \implies \mathbf{k = 5}$$ Hence, the value of $k$ for concurrency is $5$.

Common Misconceptions & Examiner Traps

Common Misconception

Neglecting to Equate Coefficients in Parallel Line Distance

Scientific Reality & Correction

First equalize the coefficients: write the first line as 6x - 8y + 18 = 0, then calculate d = |18 - (-15)| / √(6² + (-8)²) = 33/10.

Common Misconception

Allowing Negative Values for p in Normal Form

Scientific Reality & Correction

The normal form is strictly defined with p ≥ 0. If reducing Ax + By + C = 0 gives a negative constant on the right side, multiply the entire equation by -1.

Common Misconception

Applying Intercept Form x/a + y/b = 1 to Lines Passing Through Origin

Scientific Reality & Correction

A line passing through the origin has intercepts a = 0 and b = 0. Division by zero is undefined.

Common Misconception

Failing to Account for Vertical Lines in Perpendicularity

Scientific Reality & Correction

A horizontal line (m = 0) and a vertical line (m undefined) are perpendicular, but their slope product is undefined.

Common Misconception

Confusing Foot of Perpendicular and Reflection Multipliers

Scientific Reality & Correction

Foot of perpendicular requires a multiplier of -1; Image (reflection) requires a multiplier of -2.

Straight Lines Architecture, Standard Forms & Distance Formulas Diagram

Straight Lines Architecture & Standard Forms WBCHSE Class 11 Mathematics (Unit III: Coordinate Geometry) Slope & Standard Forms Slope (Gradient): m = tan θ ■ Slope-Intercept: y = mx + c ■ Point-Slope: y - y₁ = m(x - x₁) ■ Two-Point Form: y - y₁ = [(y₂ - y₁)/(x₂ - x₁)](x - x₁) ■ Intercept Form: x/a + y/b = 1 Normal (Perpendicular) Form p α x cos α + y sin α = p p = |C| / √(A² + B²) cos α = -A / ±√(A²+B²) sin α = -B / ±√(A²+B²) Length p ≥ 0 from origin, normal angle α Angle Between Lines tan θ = |(m₂ - m₁)/(1 + m₁m₂)| ■ Parallel Condition (θ = 0°): Parallel Lines: m₁ = m₂ ■ Perpendicular (θ = 90°): Perpendicular Lines: m₁ · m₂ = -1 Distance Formulas Perpendicular Distance from (x₁, y₁): d = |Ax₁ + By₁ + C| / √(A² + B²) Distance Between Parallel Lines: d = |C₁ - C₂| / √(A² + B²) Family of Lines & Reflections Family of Lines: L₁ + λL₂ = 0 ■ Foot of Perpendicular (h, k): (h - x₁)/a = (k - y₁)/b = - (ax₁ + by₁ + c)/(a² + b²) ■ Image / Reflection of Point (x', y'): (x' - x₁)/a = (y' - y₁)/b = - 2(ax₁ + by₁ + c)/(a² + b²)

Chapter Summary & 10 Key Takeaways

Takeaway 1
A straight line is the locus of a point moving with constant gradient; slope m = tan θ where 0° ≤ θ < 180° is its inclination.
Takeaway 2
Slope through two points is m = (y₂ - y₁) / (x₂ - x₁); horizontal lines have m = 0, vertical lines have undefined slope.
Takeaway 3
Two non-vertical lines are parallel if and only if m₁ = m₂, and perpendicular if and only if m₁ · m₂ = -1 (or A₁A₂ + B₁B₂ = 0).
Takeaway 4
The acute angle between two lines is given by tan θ = |(m₂ - m₁) / (1 + m₁ m₂)|.
Takeaway 5
Standard line forms include: Point-Slope y - y₁ = m(x - x₁), Slope-Intercept y = mx + c, and Intercept Form x/a + y/b = 1.
Takeaway 6
Normal form of a line is x cos α + y sin α = p, where p ≥ 0 is the perpendicular distance from origin and 0 ≤ α < 360°.
Takeaway 7
Parametric distance form is (x - x₁)/cos θ = (y - y₁)/sin θ = r, giving points (x₁ + r cos θ, y₁ + r sin θ).
Takeaway 8
Perpendicular distance from (x₁, y₁) to Ax + By + C = 0 is d = |Ax₁ + By₁ + C| / √(A² + B²).
Takeaway 9
Distance between parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 is d = |C₁ - C₂| / √(A² + B²).
Takeaway 10
Three lines are concurrent if their coefficient determinant vanishes; family of intersecting lines is L₁ + λ L₂ = 0.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
What is the slope of a line which is perpendicular to the line joining (2, 4) and (4, 8)?
Reveal Answer & Explanation
Answer: The slope of the perpendicular line is -1/2.
Find slope m₁ = (8 - 4)/(4 - 2) = 2, then use m₁ · m₂ = -1.
2
What are the x and y intercepts of the line 3x + 4y - 12 = 0?
Reveal Answer & Explanation
Answer: The intercepts are a = 4 (on x-axis) and b = 3 (on y-axis).
Convert to intercept form x/a + y/b = 1 by dividing by 12.
3
What is the perpendicular distance from the origin to the line 5x + 12y - 26 = 0?
Reveal Answer & Explanation
Answer: d = |-26| / √(25 + 144) = 26 / 13 = 2 units.
Use d = |C| / √(A² + B²) with C = -26, A = 5, B = 12.
4
What is the equation of the line parallel to 2x - 3y + 5 = 0 and passing through (1, 1)?
Reveal Answer & Explanation
Answer: 2(1) - 3(1) + k = 0 ⇒ k = 1. Equation is 2x - 3y + 1 = 0.
Write as 2x - 3y + k = 0 and substitute (1, 1) to find k.
5
What is the condition for three lines to be concurrent?
Reveal Answer & Explanation
Answer: The determinant of their coefficients must be zero: det([A1, B1, C1; A2, B2, C2; A3, B3, C3]) = 0.
Recall the 3x3 determinant of line coefficients.
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