Straight Lines — Complete Guide with Solved Examples

What Straight Lines Are Really About

Geometry in Class 11 feels like a fresh start — suddenly we’re not just drawing lines on graph paper, we’re describing them with equations. The big idea: every straight line in the coordinate plane has exactly one equation, and that equation tells us everything about the line’s slope, direction, and position.

Think of the slope as the line’s personality. A slope of 2 means “for every 1 unit you move right, move 2 units up.” A slope of −3 means you’re going steeply downhill. Once you understand slope deeply, the rest of this chapter is just variations on the same theme.

This chapter carries strong weightage in both CBSE Class 11 and JEE Main. JEE Main asks 1-2 questions from this topic almost every year, often combining it with circles or conic sections in more advanced problems. For CBSE, expect one 4-mark or 5-mark question in the board exam — and it’s the kind of question you can absolutely score full marks on with the right preparation.

Key Terms and Definitions

Slope (m): The ratio of vertical change to horizontal change between any two points on a line.

m = \frac{y_2 - y_1}{x_2 - x_1}

x-intercept: The point where the line crosses the x-axis — here, $y = 0$ .

y-intercept (c): The point where the line crosses the y-axis — here, $x = 0$ .

Collinear points: Three or more points that lie on the same straight line. The test: if the slope between any two pairs of points is equal, they’re collinear.

Angle of inclination (θ): The angle a line makes with the positive x-axis, measured counterclockwise. This means $m = \tan\theta$ , valid for $0° \leq \theta < 180°$ .

Intercept: The signed distance cut off on an axis. “Signed” matters — an x-intercept of −3 means the line crosses the x-axis to the left of the origin.

Forms of the Equation of a Line

This is the core of the chapter. There are six forms, but they’re all the same equation wearing different clothes depending on what information you’re given.

1. Slope-Intercept Form

y = mx + c

where $m$ = slope, $c$ = y-intercept

Use this when: you know slope and y-intercept directly, or when you need to quickly read off these values from an equation.

2. Point-Slope Form

y - y_1 = m(x - x_1)

where $(x_1, y_1)$ is a known point on the line

Use this when: you know the slope and one point. This is the most-used form in problem-solving because “a point and a slope” is the most common given information.

3. Two-Point Form

\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}

We derive slope automatically here. Use this when: you have two points and nothing else.

4. Intercept Form

\frac{x}{a} + \frac{y}{b} = 1

where $a$ = x-intercept, $b$ = y-intercept

Use this when: the problem gives you intercepts directly, or when you need a clean way to express the intercept condition.

The intercept form fails when the line passes through the origin (both intercepts are 0). Don’t try to write $\frac{x}{0} + \frac{y}{0} = 1$ — that’s undefined. Lines through the origin are simply $y = mx$ .

5. Normal Form

x\cos\omega + y\sin\omega = p

where $p$ = perpendicular distance from origin, $\omega$ = angle the perpendicular makes with x-axis

Use this when: you know the distance of a line from the origin and the direction of the perpendicular. Less common in JEE Main but appears in theoretical problems.

6. General Form

Ax + By + C = 0

Every straight line can be written this way. Slope $= -\frac{A}{B}$ , y-intercept $= -\frac{C}{B}$ , x-intercept $= -\frac{C}{A}$ .

Critical Formulas You Must Know

d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

Distance from point $(x_1, y_1)$ to line $Ax + By + C = 0$

\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|

Parallel: $m_1 = m_2$

Perpendicular: $m_1 \cdot m_2 = -1$

Solved Examples

Example 1 — Easy (CBSE Level)

Find the equation of a line with slope 3 and y-intercept −2.

Directly slot into $y = mx + c$ :

y = 3x + (-2) \implies y = 3x - 2

Or equivalently: $3x - y - 2 = 0$

Example 2 — Easy (CBSE Level)

Find the slope of the line joining points A(2, 5) and B(6, 1).

m = \frac{1 - 5}{6 - 2} = \frac{-4}{4} = -1

The angle of inclination: $\tan\theta = -1 \implies \theta = 135°$ .

Example 3 — Medium (CBSE / JEE Main Level)

Find the equation of the line passing through (3, −2) and perpendicular to the line $2x - 5y + 7 = 0$ .

The given line has slope $m_1 = \frac{2}{5}$ .

For the perpendicular line: $m_2 = -\frac{1}{m_1} = -\frac{5}{2}$

Using point-slope form with point (3, −2):

y - (-2) = -\frac{5}{2}(x - 3)

2(y + 2) = -5(x - 3)

2y + 4 = -5x + 15

5x + 2y - 11 = 0

Always convert to general form $Ax + By + C = 0$ in your final answer for CBSE board exams — it’s what examiners expect and avoids fraction-related marking issues.

Example 4 — Medium (JEE Main Level)

If the line $\frac{x}{a} + \frac{y}{b} = 1$ passes through (1, 2) and (3, 4), find a and b.

Substituting both points:

\frac{1}{a} + \frac{2}{b} = 1 \quad \text{...(i)}

\frac{3}{a} + \frac{4}{b} = 1 \quad \text{...(ii)}

Let $u = \frac{1}{a}$ and $v = \frac{1}{b}$ :

From (i): $u + 2v = 1$ From (ii): $3u + 4v = 1$

Multiply (i) by 2: $2u + 4v = 2$

Subtract from (ii): $u = -1 \implies a = -1$

Substitute back: $-1 + 2v = 1 \implies v = 1 \implies b = 1$

So $a = -1, b = 1$ — meaning the line crosses x-axis at −1 and y-axis at 1.

Example 5 — Hard (JEE Main Level)

Find the distance between the parallel lines $3x - 4y + 7 = 0$ and $3x - 4y - 5 = 0$ .

For parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ , the distance formula is:

d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

d = \frac{|7 - (-5)|}{\sqrt{9 + 16}} = \frac{12}{\sqrt{25}} = \frac{12}{5} = 2.4 \text{ units}

This exact formula appeared in JEE Main 2024 January session. The trick is to first confirm the lines are actually parallel (same $A$ and $B$ coefficients) before applying this. If coefficients aren’t equal but proportional, rewrite them to match first.

Exam-Specific Tips

CBSE Board (Class 11 and 12)

CBSE typically asks for equation derivation (2-mark or 3-mark), distance from a point to a line (3-mark), and occasionally angle between lines (4-mark). The marking scheme awards partial marks, so always write the formula before substituting — even a correct numerical answer without the formula shown can cost you 1 mark.

For the board exam, memorise all six forms. The examiner’s choice of form depends on what data is given, and using the “wrong” (though technically correct) form can make your working unnecessarily lengthy.

JEE Main

JEE Main asks 1-2 questions from straight lines, often as single-answer MCQs. Common patterns:

Finding equations of lines satisfying two conditions simultaneously
Distance/locus problems involving families of lines
Combining with the distance formula for triangle area

The family of lines concept is a JEE Main favourite: if $L_1 = 0$ and $L_2 = 0$ are two lines, then $L_1 + \lambda L_2 = 0$ represents the family of all lines through their intersection. This lets you find specific lines without finding the intersection point first — saves 30 seconds on exam day.

SAT / International

SAT tests slope and intercept in the context of linear models and word problems. The equation is almost always in $y = mx + b$ form. Focus on interpreting $m$ and $b$ in real-world contexts (“rate of change” and “initial value”).

Common Mistakes to Avoid

Mistake 1: Forgetting the absolute value in distance formulas. The distance from a point to a line is always positive. Write $\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2+B^2}}$ , not without the bars. Without the modulus, you’ll get a negative answer for some points — and lose marks.

Mistake 2: Saying perpendicular slopes “multiply to give −1” when one slope is undefined. A vertical line (slope undefined, equation $x = k$ ) is perpendicular to a horizontal line (slope 0, equation $y = k$ ). The formula $m_1 m_2 = -1$ breaks down here. State this case separately.

Mistake 3: Confusing angle of inclination with angle between two lines. $\tan\theta = m$ gives the angle with the positive x-axis (inclination). The angle between two lines uses the formula with $m_1$ and $m_2$ — different formula entirely.

Mistake 4: Not checking if three points are collinear before using the two-point form for all three. If a problem says “find the line through A, B, C”, first verify they’re collinear using the slope test. If they’re not, you can’t write a single line through all three — and the problem might be asking you to identify this.

Mistake 5: Applying intercept form to lines through the origin. If the line passes through $(0,0)$ , both intercepts are 0. The intercept form $\frac{x}{a} + \frac{y}{b} = 1$ doesn’t work. Just use $y = mx$ directly.

Practice Questions

Q1. Find the slope and y-intercept of the line $4x - 3y + 12 = 0$ .

Rearrange: $3y = 4x + 12 \implies y = \frac{4}{3}x + 4$

Slope $= \frac{4}{3}$ , y-intercept $= 4$

Q2. Find the equation of the line passing through $(-1, 4)$ and parallel to $3x + 5y = 7$ .

Slope of given line $= -\frac{3}{5}$ . Parallel line has the same slope.

Using point-slope: $y - 4 = -\frac{3}{5}(x + 1)$

$5y - 20 = -3x - 3 \implies 3x + 5y - 17 = 0$

Q3. Find the area of the triangle formed by the line $2x + 3y = 6$ with the coordinate axes.

x-intercept: set $y=0$ : $x = 3$ . So point is $(3, 0)$ .

y-intercept: set $x=0$ : $y = 2$ . So point is $(0, 2)$ .

Area $= \frac{1}{2} \times base \times height = \frac{1}{2} \times 3 \times 2 = 3$ sq. units

Q4. Show that the points $(1, 1)$ , $(2, 3)$ , and $(3, 5)$ are collinear.

Slope between $(1,1)$ and $(2,3)$ : $\frac{3-1}{2-1} = 2$

Slope between $(2,3)$ and $(3,5)$ : $\frac{5-3}{3-2} = 2$

Both slopes equal $\implies$ the three points are collinear.

Q5. Find the distance of the point $(3, -4)$ from the line $12x - 5y + 2 = 0$ .

d = \frac{|12(3) - 5(-4) + 2|}{\sqrt{144 + 25}} = \frac{|36 + 20 + 2|}{\sqrt{169}} = \frac{58}{13}

Q6. Find the angle between lines $y = \sqrt{3}x + 1$ and $y = \frac{1}{\sqrt{3}}x - 2$ .

$m_1 = \sqrt{3}$ , $m_2 = \frac{1}{\sqrt{3}}$

\tan\theta = \left|\frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} \cdot \frac{1}{\sqrt{3}}}\right| = \left|\frac{\frac{3-1}{\sqrt{3}}}{1+1}\right| = \left|\frac{2}{2\sqrt{3}}\right| = \frac{1}{\sqrt{3}}

$\theta = 30°$

Q7. Find $k$ if the lines $3x - ky - 8 = 0$ and $6x - 8y + 1 = 0$ are parallel.

For parallel lines: $\frac{A_1}{A_2} = \frac{B_1}{B_2}$ (but $\neq \frac{C_1}{C_2}$ )

\frac{3}{6} = \frac{-k}{-8} \implies \frac{1}{2} = \frac{k}{8} \implies k = 4

Verify: $\frac{C_1}{C_2} = \frac{-8}{1} \neq \frac{1}{2}$ ✓ (so they’re parallel, not the same line)

Q8 (JEE-level). Find the equation of the line through the intersection of $x + 2y = 3$ and $3x - y = 5$ , and perpendicular to $2x - 3y + 4 = 0$ .

Using family of lines: $(x + 2y - 3) + \lambda(3x - y - 5) = 0$

$(1 + 3\lambda)x + (2 - \lambda)y + (-3 - 5\lambda) = 0$

Slope of this line: $-\frac{1+3\lambda}{2-\lambda}$

Slope of $2x - 3y + 4 = 0$ is $\frac{2}{3}$ .

For perpendicular: $\left(-\frac{1+3\lambda}{2-\lambda}\right) \cdot \frac{2}{3} = -1$

$\frac{2(1+3\lambda)}{3(2-\lambda)} = 1 \implies 2 + 6\lambda = 6 - 3\lambda \implies 9\lambda = 4 \implies \lambda = \frac{4}{9}$

Substituting: $(1 + \frac{12}{9})x + (2 - \frac{4}{9})y - (3 + \frac{20}{9}) = 0$

$\frac{21}{9}x + \frac{14}{9}y - \frac{47}{9} = 0 \implies 21x + 14y - 47 = 0$

Frequently Asked Questions

What is the equation of the x-axis and y-axis?

The x-axis has equation $y = 0$ (slope = 0, passes through origin along horizontal). The y-axis has equation $x = 0$ — it’s a vertical line with undefined slope.

How do I find whether two lines are parallel, perpendicular, or neither?

Compare slopes: $m_1 = m_2$ → parallel; $m_1 m_2 = -1$ → perpendicular; anything else → they intersect at an angle. For lines in general form $Ax + By + C = 0$ : compare $\frac{A_1}{A_2}$ and $\frac{B_1}{B_2}$ — if equal, parallel; if $A_1 A_2 + B_1 B_2 = 0$ , perpendicular.

What is the locus of a point equidistant from two given lines?

This gives the angle bisectors of the two lines. If the lines are $L_1 = 0$ and $L_2 = 0$ , the bisectors are given by $\frac{L_1}{\sqrt{A_1^2+B_1^2}} = \pm\frac{L_2}{\sqrt{A_2^2+B_2^2}}$ . This is a common JEE topic.

How many equations can a straight line have?

Infinitely many — any scalar multiple of $Ax + By + C = 0$ gives the same line. But the line’s geometric identity is unique. We typically write the equation with integers and no common factor.

What does a negative slope mean?

The line falls from left to right. As x increases, y decreases. The steeper the negative slope (more negative), the faster the fall.

How do I find the foot of the perpendicular from a point to a line?

Drop a perpendicular from the point to the line. The foot is found by: writing the perpendicular line through the given point, then solving the two-line system. There’s also a direct formula — for point $(x_1, y_1)$ and line $Ax + By + C = 0$ , the foot $(h, k)$ satisfies $\frac{h - x_1}{A} = \frac{k - y_1}{B} = -\frac{Ax_1 + By_1 + C}{A^2 + B^2}$ .

Why does the distance formula have an absolute value?

Because $Ax_1 + By_1 + C$ can be positive or negative depending on which side of the line the point lies. Distance is always a positive quantity, so we take the absolute value.

What is the significance of the general form $Ax + By + C = 0$ ?

It’s the most flexible — every line (including vertical ones, which can’t be written as $y = mx + c$ ) fits this form. In JEE, converting all lines to general form before comparing or combining them avoids slope-undefined errors with vertical lines.