Intermediate

28 min read

Straight Lines — Slope, Forms of Equations & Intercepts

Master Straight Lines for JEE Main JEE Advanced and Board Exams. Covers slope gradient parallel perpendicular conditions all standard forms of line equations distance from point to line parallel lines distance foot of perpendicular reflection concurrent lines locus problems with fully verified solved examples.

New

The straight line is the simplest curve in coordinate geometry — yet its study reveals a remarkably rich structure. Every straight line can be described in multiple equivalent forms, each useful in different contexts. Understanding the relationships between slopes, intercepts, and angles is the foundation for the entire chapter of coordinate geometry, and it appears as a prerequisite in conic sections, 3D geometry, and calculus (tangents and normals). For JEE (Main & Advanced), straight lines contribute 2–3 direct questions per paper — ranging from finding equations of lines and distances from points to problems on concurrent lines and locus. A clean, systematic approach to this topic prevents the most common algebraic errors.

1. Basic Concepts — Distance, Section, Slope

Distance Formula

Distance between two points $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ :

$A B = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

Section Formula

Point dividing $A B$ in ratio $m : n$ :

Internal division: $P = (\frac{m x_{2} + n x_{1}}{m + n}, \frac{m y_{2} + n y_{1}}{m + n})$
External division: $P = (\frac{m x_{2} - n x_{1}}{m - n}, \frac{m y_{2} - n y_{1}}{m - n})$
Midpoint ( $m = n = 1$ ): $M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$

Centroid, Incentre, and Circumcentre of a Triangle

Point	Coordinates	Property
Centroid (G)	$(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$	Intersection of medians; divides each median in $2 : 1$
Incentre (I)	$(\frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}, \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c})$	Intersection of angle bisectors; $a, b, c$ are side lengths opposite to $A, B, C$
Area of $△ A B C$	$\frac{1}{2} \| x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) \|$	= 0 iff points are collinear

2. Slope (Gradient) of a Line

The slope (or gradient) of a line is the tangent of the angle $α$ it makes with the positive $x$ -axis (measured anticlockwise):

$m = \tan α$ ( $α \neq 90 °$ )

For a line through $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ ( $x_{1} \neq x_{2}$ )

Condition	Slope	Line orientation
$m > 0$	Positive	Rises left to right (acute angle with $x$ -axis)
$m < 0$	Negative	Falls left to right (obtuse angle with $x$ -axis)
$m = 0$	Zero	Horizontal line (parallel to $x$ -axis)
$m$ undefined	$\tan 90 °$ = undefined	Vertical line (parallel to $y$ -axis)
$m_{1} = m_{2}$	Equal slopes	Lines are parallel (or identical)
$m_{1} \cdot m_{2} = - 1$	Product $= - 1$	Lines are perpendicular

Angle Between Two Lines

The acute angle $θ$ between two lines with slopes $m_{1}$ and $m_{2}$ :

$\tan θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} |$

If $1 + m_{1} m_{2} = 0$ : lines are perpendicular. If $m_{1} = m_{2}$ : lines are parallel ( $θ = 0$ ).

3. Standard Forms of a Line

Form	Equation	Given	Note
Slope-Intercept	$y = m x + c$	Slope $m$ , $y$ -intercept $c$	Most commonly used form
Point-Slope	$y - y_{1} = m (x - x_{1})$	Slope $m$ , point $(x_{1}, y_{1})$	Use when point and slope given
Two-Point	$\frac{y - y_{1}}{y_{2} - y_{1}} = \frac{x - x_{1}}{x_{2} - x_{1}}$	Two points $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$	Equivalent to point-slope with $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$
Intercept	$\frac{x}{a} + \frac{y}{b} = 1$	$x$ -intercept $a$ , $y$ -intercept $b$	Fails for lines through origin or parallel to axes
Normal (Perpendicular)	$x \cos α + y \sin α = p$	$p$ = perpendicular distance from origin; $α$ = angle normal makes with $x$ -axis	$p > 0$ always; $0 \leq α < 2 π$
General	$a x + b y + c = 0$	Coefficients $a, b, c$	Slope $= - a / b$ ; $x$ -int $= - c / a$ ; $y$ -int $= - c / b$

Key Properties from General Form $a x + b y + c = 0$

Slope: $m = - \frac{a}{b}$
$x$ -intercept (set $y = 0$ ): $x = - \frac{c}{a}$
$y$ -intercept (set $x = 0$ ): $y = - \frac{c}{b}$
Parallel line through $(x_{1}, y_{1})$ : $a (x - x_{1}) + b (y - y_{1}) = 0$ , i.e., $a x + b y = a x_{1} + b y_{1}$
Perpendicular line through $(x_{1}, y_{1})$ : $b (x - x_{1}) - a (y - y_{1}) = 0$ , i.e., $b x - a y = b x_{1} - a y_{1}$

4. Distance Formulae

Distance of a Point from a Line

Perpendicular distance from point $(x_{1}, y_{1})$ to line $a x + b y + c = 0$ :

$d = \frac{| a x_{1} + b y_{1} + c |}{\sqrt{a^{2} + b^{2}}}$

Distance Between Two Parallel Lines

For parallel lines $a x + b y + c_{1} = 0$ and $a x + b y + c_{2} = 0$ :

$d = \frac{| c_{1} - c_{2} |}{\sqrt{a^{2} + b^{2}}}$

Foot of Perpendicular from a Point to a Line

Foot of perpendicular from $(x_{1}, y_{1})$ to $a x + b y + c = 0$ :

$\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = - \frac{(a x_{1} + b y_{1} + c)}{a^{2} + b^{2}}$

Reflection (Image) of a Point in a Line

Image of $(x_{1}, y_{1})$ in line $a x + b y + c = 0$ :

$\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = - \frac{2 (a x_{1} + b y_{1} + c)}{a^{2} + b^{2}}$

Note: The foot of perpendicular formula uses factor $- 1$ ; the reflection formula uses factor $- 2$ . The image point is twice as far as the foot from the original point, on the other side of the line.

5. Position of a Point Relative to a Line

For the line $L : a x + b y + c = 0$ , substitute the point coordinates into $a x + b y + c$ :

Points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are on the same side of $L$ if $(a x_{1} + b y_{1} + c)$ and $(a x_{2} + b y_{2} + c)$ have the same sign.
They are on opposite sides if the expressions have opposite signs.
The origin is on the side where $c$ and $a x + b y + c$ have the same sign (just substitute $x = y = 0$ giving $c$ ).

6. Concurrent Lines

Three lines $L_{1}$ , $L_{2}$ , $L_{3}$ are concurrent (pass through a common point) if and only if the determinant of their coefficients is zero:

For $a_{1} x + b_{1} y + c_{1} = 0$ , $a_{2} x + b_{2} y + c_{2} = 0$ , $a_{3} x + b_{3} y + c_{3} = 0$ :

$| \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} | = 0$

Alternative: Lines of the form $L_{1} + λ L_{2} = 0$

Any line passing through the intersection of $L_{1} = 0$ and $L_{2} = 0$ can be written as $L_{1} + λ L_{2} = 0$ for some constant $λ$ . Use an additional condition to determine $λ$ .

7. Locus Problems

A locus is the set of all points satisfying a given geometric condition. To find the equation of a locus:

Step 1: Let the moving point be $P (h, k)$ .
Step 2: Write the geometric condition as an equation in $h$ and $k$ .
Step 3: Eliminate any parameters (if present) using the constraints.
Step 4: Replace $h \to x$ and $k \to y$ to get the locus equation.

Worked Example

Find the locus of a point $P$ such that the sum of its distances from $(3, 0)$ and $(- 3, 0)$ equals $8$ .

Let $P = (h, k)$ . Condition: $\sqrt{(h - 3)^{2} + k^{2}} + \sqrt{(h + 3)^{2} + k^{2}} = 8$ .
Isolate one radical: $\sqrt{(h - 3)^{2} + k^{2}} = 8 - \sqrt{(h + 3)^{2} + k^{2}}$ .
Squaring: $(h - 3)^{2} + k^{2} = 64 - 16 \sqrt{(h + 3)^{2} + k^{2}} + (h + 3)^{2} + k^{2}$ .
Simplify: $- 12 h - 64 = - 16 \sqrt{(h + 3)^{2} + k^{2}}$ → $3 h + 16 = 4 \sqrt{(h + 3)^{2} + k^{2}}$ .
Square again: $9 h^{2} + 96 h + 256 = 16 (h^{2} + 6 h + 9 + k^{2})$ → $7 h^{2} + 16 k^{2} = 112$ .
Locus: $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$ (an ellipse).

Straight Lines — Speed Framework for JEE

Problem Type	Go-to Approach
Equation through two points	Two-point form; or find $m$ first, then point-slope
Line parallel to $a x + b y + c = 0$	Same $a, b$ coefficients: $a x + b y + k = 0$ ; find $k$ from condition
Line perpendicular to $a x + b y + c = 0$	Swap and negate: $b x - a y + k = 0$ ; find $k$ from condition
Distance from point to line	$d = \| a x_{1} + b y_{1} + c \| / \sqrt{a^{2} + b^{2}}$ ; always positive
Three lines concurrent	$3 \times 3$ determinant = 0, OR solve two lines and check in third
Line through intersection of $L_{1}, L_{2}$	Write as $L_{1} + λ L_{2} = 0$ ; use extra condition for $λ$
Image of point in line	Use reflection formula; or find foot, then double the displacement

JEE Power Tips — Avoid These Common Mistakes

Perpendicular lines: product of slopes $= - 1$ , not just "swap slopes." If $m_{1} = 2$ , then $m_{2} = - \frac{1}{2}$ , not $- 2$ . Many students negate without taking the reciprocal.
Parallel lines $a x + b y + c_{1} = 0$ and $a x + b y + c_{2} = 0$ — distance $= | c_{1} - c_{2} | / \sqrt{a^{2} + b^{2}}$ . The lines must be in the same form (same $a$ and $b$ coefficients) before applying this formula. Always verify they are truly parallel first ( $m_{1} = m_{2}$ ).
The intercept form $x / a + y / b = 1$ fails for lines through the origin ( $c = 0$ ) and for horizontal/vertical lines. Never force a line into intercept form without checking these conditions.
Area of triangle formula sign: The signed area can be negative — always take the absolute value. The area is zero iff the three points are collinear.
Angle between lines formula gives the acute angle — use the absolute value of the tangent expression. The obtuse angle would be $180 ° - θ$ .
$L_{1} + λ L_{2} = 0$ represents all lines through the intersection of $L_{1}$ and $L_{2}$ except $L_{2}$ itself. To get $L_{2}$ , take $λ \to \infty$ (write as $μ L_{1} + L_{2} = 0$ and let $μ \to 0$ ).
Foot of perpendicular uses $- 1$ ; reflection uses $- 2$ in the parameter formula. Confusing these gives the foot when you need the reflection and vice versa — a very common error in JEE.

Practice Questions (JEE / Board Level)

Q1: The slope of the line passing through $(2, 3)$ and $(- 4, 1)$ is:

A) $\frac{1}{3}$
B) $\frac{1}{2}$
C) 3
D) 2

Answer: A) $\frac{1}{3}$ .

Explanation:

The slope $m$ of a line passing through two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by:
$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

Substitute the given points $(2, 3)$ and $(- 4, 1)$ :
$m = \frac{1 - 3}{- 4 - 2}$
$m = \frac{- 2}{- 6} = \frac{1}{3}$ .

Q2: The equation of the line with $x$ -intercept 3 and $y$ -intercept -4 is:

A) $4 x - 3 y = 12$
B) $3 x - 4 y = 12$
C) $4 x + 3 y = 12$
D) $3 x + 4 y = 12$

Answer: A) $4 x - 3 y = 12$ .

Explanation:

Use the intercept form of a line equation, $\frac{x}{a} + \frac{y}{b} = 1$ , where $a$ and $b$ are the $x$ and $y$ intercepts respectively.

Substitute $a = 3$ and $b = - 4$ :
$\frac{x}{3} + \frac{y}{- 4} = 1$

Multiply the entire equation by the least common multiple (12) to clear the fractions:
$12 \cdot (\frac{x}{3} - \frac{y}{4}) = 12 \cdot 1$
$4 x - 3 y = 12$ .

Q3: The perpendicular distance from the point $(1, 3)$ to the line $4 x - 3 y + 1 = 0$ is:

A) $\frac{4}{5}$
B) $\frac{3}{5}$
C) $\frac{6}{5}$
D) $\frac{8}{5}$

Answer: A) $\frac{4}{5}$ .

Explanation:

The perpendicular distance $d$ from a point $(x_{1}, y_{1})$ to a line $a x + b y + c = 0$ is given by:
$d = \frac{| a x_{1} + b y_{1} + c |}{\sqrt{a^{2} + b^{2}}}$

Substitute $(x_{1}, y_{1}) = (1, 3)$ and the line coefficients $a = 4, b = - 3, c = 1$ :
$d = \frac{| 4 (1) - 3 (3) + 1 |}{\sqrt{4^{2} + (- 3)^{2}}}$
$d = \frac{| 4 - 9 + 1 |}{\sqrt{16 + 9}}$
$d = \frac{| - 4 |}{\sqrt{25}} = \frac{4}{5}$ .

Q4: The lines $x + 2 y - 4 = 0$ , $x - y - 1 = 0$ , and $3 x + k y - 6 = 0$ are concurrent. The value of $k$ is:

A) 0
B) 2
C) 4
D) -2

Answer: A) 0.

Explanation:

Since the lines are concurrent, they all intersect at exactly one common point. First, find the intersection of the first two lines:
1) $x + 2 y = 4$
2) $x - y = 1$

Subtract equation (2) from equation (1):
$(x + 2 y) - (x - y) = 4 - 1$
$3 y = 3 ⟹ y = 1$ .

Substitute $y = 1$ into equation (2):
$x - 1 = 1 ⟹ x = 2$ .
The point of concurrency is $(2, 1)$ .

Substitute this point into the third line's equation to find $k$ :
$3 (2) + k (1) - 6 = 0$
$6 + k - 6 = 0 ⟹ k = 0$ .

Q5: If the lines $2 x - k y = 3$ and $3 x + 2 y = 5$ are perpendicular to each other, the value of $k$ is:

A) -3
B) 3
C) $\frac{1}{3}$
D) $- \frac{1}{3}$

Answer: B) 3.

Explanation:

Find the slopes of both lines by converting them to slope-intercept form ( $y = m x + c$ ).

First line: $2 x - k y = 3 ⟹ k y = 2 x - 3 ⟹ y = \frac{2}{k} x - \frac{3}{k}$ .
Slope $m_{1} = \frac{2}{k}$ .

Second line: $3 x + 2 y = 5 ⟹ 2 y = - 3 x + 5 ⟹ y = - \frac{3}{2} x + \frac{5}{2}$ .
Slope $m_{2} = - \frac{3}{2}$ .

For two lines to be perpendicular, the product of their slopes must be -1 ( $m_{1} \cdot m_{2} = - 1$ ):
$(\frac{2}{k}) \cdot (- \frac{3}{2}) = - 1$
$- \frac{3}{k} = - 1 ⟹ k = 3$ .

Q6: The distance between the parallel lines $3 x + 4 y - 5 = 0$ and $6 x + 8 y - 15 = 0$ is:

A) $\frac{1}{2}$
B) $\frac{1}{10}$
C) $\frac{5}{2}$
D) $\frac{1}{5}$

Answer: A) $\frac{1}{2}$ .

Explanation:

To use the parallel distance formula, both line equations must have identical $x$ and $y$ coefficients. Multiply the first equation by 2:
$2 \cdot (3 x + 4 y - 5 = 0) ⟹ 6 x + 8 y - 10 = 0$ .

Now we have two parallel lines:
Line 1: $6 x + 8 y - 10 = 0$ (so $c_{1} = - 10$ )
Line 2: $6 x + 8 y - 15 = 0$ (so $c_{2} = - 15$ )

Apply the distance formula $d = \frac{| c_{1} - c_{2} |}{\sqrt{a^{2} + b^{2}}}$ :
$d = \frac{| - 10 - (- 15) |}{\sqrt{6^{2} + 8^{2}}}$
$d = \frac{| - 10 + 15 |}{\sqrt{36 + 64}} = \frac{5}{\sqrt{100}} = \frac{5}{10} = \frac{1}{2}$ .

Q7 (JEE type): A line is drawn through the point $(1, 2)$ to meet the coordinate axes at $P$ and $Q$ . If the area of the triangle $O P Q$ is minimum, the equation of the line is:

A) $x + 2 y = 5$
B) $2 x + y = 4$
C) $x + 2 y = 6$
D) $2 x + y = 6$

Answer: B) $2 x + y = 4$ .

Explanation:

Let the equation of the line in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$ .
The line meets the axes at $P (a, 0)$ and $Q (0, b)$ . The area of the right-angled triangle $O P Q$ is $Area = \frac{1}{2} | a b |$ .

Since the point $(1, 2)$ lies on the line, it must satisfy the equation:
$\frac{1}{a} + \frac{2}{b} = 1$

To minimize the area (which means minimizing $| a b |$ ), apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality to the positive terms $\frac{1}{a}$ and $\frac{2}{b}$ :
$\frac{\frac{1}{a} + \frac{2}{b}}{2} \geq \sqrt{\frac{1}{a} \cdot \frac{2}{b}}$

Substitute the constraint $\frac{1}{a} + \frac{2}{b} = 1$ :
$\frac{1}{2} \geq \sqrt{\frac{2}{a b}}$

Square both sides:
$\frac{1}{4} \geq \frac{2}{a b} ⟹ a b \geq 8$ .
Therefore, the minimum area is $\frac{1}{2} (8) = 4$ .

The minimum occurs when the equality condition of AM-GM is met, meaning the two terms must be equal:
$\frac{1}{a} = \frac{2}{b} ⟹ b = 2 a$ .

Substitute $b = 2 a$ back into the constraint equation:
$\frac{1}{a} + \frac{2}{2 a} = 1$
$\frac{2}{a} = 1 ⟹ a = 2$ .
So, $b = 2 (2) = 4$ .

The equation of the line is:
$\frac{x}{2} + \frac{y}{4} = 1 ⟹ 2 x + y = 4$ .

Related Study Material

More Mathematics Notes More Coordinate Geometry Notes All Straight Lines Topics

Practice Questions

Mathematics Practice Straight Lines Questions

Topic Information

DifficultyIntermediate

Est. Read Time28 minutes

CategoryStraight Lines

Quick Actions

Track Your Learning

Study Tips

Take notes of key points while reading

Practice related questions after studying

Review topics marked as "Learning" regularly

Test your knowledge with practice questions

Edvaya Target

Detailed Notes & Solved Examples

Master complex concepts with our rich, topic-wise study notes. Access hundreds of step-by-step solved examples and interactive flashcards right from your phone.

Edvaya Target

Edvaya Aspire

Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Straight Lines — Slope, Forms of Equations & Intercepts

1. Basic Concepts — Distance, Section, Slope

Distance Formula

Section Formula

Centroid, Incentre, and Circumcentre of a Triangle

2. Slope (Gradient) of a Line

Angle Between Two Lines

3. Standard Forms of a Line

Key Properties from General Form $a x + b y + c = 0$

4. Distance Formulae

Distance of a Point from a Line

Distance Between Two Parallel Lines

Foot of Perpendicular from a Point to a Line

Reflection (Image) of a Point in a Line

5. Position of a Point Relative to a Line

6. Concurrent Lines

Alternative: Lines of the form $L_{1} + λ L_{2} = 0$

7. Locus Problems

Worked Example

Straight Lines — Speed Framework for JEE

Practice Questions (JEE / Board Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

Edvaya Target

We Value Your Privacy

Edvaya Target

Edvaya Aspire

Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Straight Lines — Slope, Forms of Equations & Intercepts

1. Basic Concepts — Distance, Section, Slope

Distance Formula

Section Formula

Centroid, Incentre, and Circumcentre of a Triangle

2. Slope (Gradient) of a Line

Angle Between Two Lines

3. Standard Forms of a Line

Key Properties from General Form ax+by+c=0

4. Distance Formulae

Distance of a Point from a Line

Distance Between Two Parallel Lines

Foot of Perpendicular from a Point to a Line

Reflection (Image) of a Point in a Line

5. Position of a Point Relative to a Line

6. Concurrent Lines

Alternative: Lines of the form L1+λL2=0

7. Locus Problems

Worked Example

Straight Lines — Speed Framework for JEE

Practice Questions (JEE / Board Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

Edvaya Target

We Value Your Privacy

Key Properties from General Form $a x + b y + c = 0$

Alternative: Lines of the form $L_{1} + λ L_{2} = 0$