The Hook: When Your GPS Says “Continue Straight”
Ever wondered how your GPS navigates you through the city? When it says “Continue straight for 2.5 km then turn left at 45 degrees,” it’s using coordinate geometry!
- Your phone’s GPS uses coordinates (latitude, longitude)
- It calculates the shortest distance between points
- It finds angles between roads to give you turn-by-turn directions
The Question: How do mathematicians describe “straight” precisely? How do we find angles between roads or the shortest distance from your location to a highway?
JEE Impact: This topic appears in 3-4 questions every year in JEE Main and 1-2 in Advanced. Master this, and coordinate geometry becomes easy!
The Core Concept
A straight line is the shortest path between two points. In coordinate geometry, we describe this path using equations.
The Big Idea
Think of a straight line as an infinite collection of points $(x, y)$ that satisfy a certain relationship. This relationship can be expressed in different forms depending on what information we have:
- If we know slope and a point: Use point-slope form
- If we know two points: Use two-point form
- If we know intercepts: Use intercept form
- For general cases: Use slope-intercept or general form
All Forms of Straight Line Equations
1. Slope-Intercept Form
$$\boxed{y = mx + c}$$Interactive Demo: Visualize Straight Lines
Plot and explore straight lines with different slopes and intercepts.
In simple terms: “The height $y$ depends on how steep the line is ($m$) and where it starts on the y-axis ($c$)”
- $m$ = slope = “steepness” or “gradient” of the line
- $c$ = y-intercept = where the line crosses the y-axis
When to use: When you know the slope and y-intercept directly.
A car travels such that for every 3 km eastward (x-direction), it goes 4 km northward (y-direction), starting from 5 km north of origin.
Slope $m = \frac{4}{3}$, y-intercept $c = 5$
Equation: $y = \frac{4}{3}x + 5$
2. Point-Slope Form
$$\boxed{y - y_1 = m(x - x_1)}$$In simple terms: “Starting from point $(x_1, y_1)$, for every step in x-direction, we move $m$ steps in y-direction”
When to use: When you know one point and the slope.
3. Two-Point Form
$$\boxed{y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)}$$Or more symmetrically:
$$\boxed{\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}}$$In simple terms: “If the line passes through two points, we can find its direction from those points”
When to use: When you know two points on the line.
4. Intercept Form
$$\boxed{\frac{x}{a} + \frac{y}{b} = 1}$$In simple terms: “The line cuts the x-axis at $(a, 0)$ and y-axis at $(0, b)$”
- $a$ = x-intercept
- $b$ = y-intercept
When to use: When you know both intercepts (and neither is zero).
5. Normal Form (Perpendicular Form)
$$\boxed{x \cos \alpha + y \sin \alpha = p}$$In simple terms: “Draw a perpendicular from origin to the line. If this perpendicular has length $p$ and makes angle $\alpha$ with x-axis, this is your equation”
- $p$ = perpendicular distance from origin to line
- $\alpha$ = angle the perpendicular makes with positive x-axis
When to use: Rarely in JEE, but useful for finding distance from origin.
6. General Form
$$\boxed{Ax + By + C = 0}$$In simple terms: “The most general way to write any straight line”
Key Relations:
- Slope: $m = -\frac{A}{B}$ (if $B \neq 0$)
- x-intercept: $-\frac{C}{A}$ (if $A \neq 0$)
- y-intercept: $-\frac{C}{B}$ (if $B \neq 0$)
Slope: The Measure of Steepness
Definition
$$\boxed{m = \tan \theta = \frac{y_2 - y_1}{x_2 - x_1}}$$where $\theta$ is the angle the line makes with the positive x-axis.
Understanding Slope
| Slope | Meaning | Example |
|---|---|---|
| $m > 0$ | Line rises as we move right | Uphill road |
| $m < 0$ | Line falls as we move right | Downhill road |
| $m = 0$ | Horizontal line | Flat road parallel to ground |
| $m = \infty$ | Vertical line | Wall perpendicular to ground |
“RISE over RUN”
$$m = \frac{\text{RISE (change in y)}}{\text{RUN (change in x)}} = \frac{\Delta y}{\Delta x}$$Think of climbing stairs: RISE = vertical height, RUN = horizontal distance.
Angle Between Two Lines
Formula
If two lines have slopes $m_1$ and $m_2$, the acute angle $\theta$ between them is:
$$\boxed{\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|}$$Visual Description: Imagine two roads crossing. The angle between them is measured by the tangent formula above.
Special Cases
1. Parallel Lines
$$\boxed{m_1 = m_2 \quad \text{or} \quad \frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2}}$$In simple terms: “Same slope, different intercepts” → Lines never meet (parallel tracks)
2. Perpendicular Lines
$$\boxed{m_1 \cdot m_2 = -1 \quad \text{or} \quad A_1 A_2 + B_1 B_2 = 0}$$In simple terms: “Product of slopes is -1” → Lines meet at 90° (like adjacent walls)
Are these lines perpendicular?
Line 1: $2x + 3y = 5$ → slope $m_1 = -\frac{2}{3}$
Line 2: $3x - 2y = 7$ → slope $m_2 = \frac{3}{2}$
Check: $m_1 \cdot m_2 = -\frac{2}{3} \times \frac{3}{2} = -1$ ✓
Yes, they’re perpendicular!
3. Coincident Lines
$$\boxed{\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}}$$In simple terms: “Same line written differently”
Distance Formulas
1. Distance from Point to Line
The perpendicular distance from point $(x_1, y_1)$ to line $Ax + By + C = 0$ is:
$$\boxed{d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}}$$Memory Trick: “Add Best Coordinates, Divide by Root”
Visual Description: Imagine dropping a perpendicular from a point to a road. This formula gives that shortest distance.
Find distance from point $(1, 2)$ to line $3x + 4y - 5 = 0$
$$d = \frac{|3(1) + 4(2) - 5|}{\sqrt{3^2 + 4^2}} = \frac{|3 + 8 - 5|}{\sqrt{9 + 16}} = \frac{6}{5} \text{ units}$$2. Distance Between Parallel Lines
For parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$:
$$\boxed{d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}}$$Visual Description: Width of a road with parallel edges, measured perpendicular to the edges.
3. Distance Between Two Points
$$\boxed{d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}$$This is just the Pythagorean theorem!
Position of Point Relative to Line
For a line $Ax + By + C = 0$, substitute point $(x_1, y_1)$:
- If $Ax_1 + By_1 + C = 0$ → Point is on the line
- If $Ax_1 + By_1 + C > 0$ → Point is on one side
- If $Ax_1 + By_1 + C < 0$ → Point is on the other side
Application: Two points $(x_1, y_1)$ and $(x_2, y_2)$ are on:
- Same side if $Ax_1 + By_1 + C$ and $Ax_2 + By_2 + C$ have the same sign
- Opposite sides if they have opposite signs
Memory Tricks & Patterns
Mnemonic for Forms
"Point-Slope Takes Me Inside New Galaxy"
- Point-Slope: $y - y_1 = m(x - x_1)$
- Slope-Intercept: $y = mx + c$
- Two-Point: Uses two points
- M→ Slope symbol
- Intercept: $\frac{x}{a} + \frac{y}{b} = 1$
- Normal: $x\cos\alpha + y\sin\alpha = p$
- General: $Ax + By + C = 0$
Pattern Recognition
90% of JEE distance problems use this formula:
$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$Pro Tip: First convert any line equation to $Ax + By + C = 0$ form for easy distance calculation.
Slope Relationships
| Relationship | Slope Condition | Visual |
|---|---|---|
| Parallel | $m_1 = m_2$ | Railway tracks |
| Perpendicular | $m_1 m_2 = -1$ | Corner of a room |
| $45°$ angle | $m = 1$ | Diagonal of a square |
| $135°$ angle | $m = -1$ | Other diagonal |
When to Use Which Form
Given Information → Form to Use
- Slope + y-intercept → Use $y = mx + c$
- Slope + one point → Use $y - y_1 = m(x - x_1)$
- Two points → Use $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$
- Both intercepts → Use $\frac{x}{a} + \frac{y}{b} = 1$
- For calculations → Convert to $Ax + By + C = 0$
Exam Strategy: In JEE, always convert to general form $Ax + By + C = 0$ for final answer unless specifically asked otherwise.
Common Mistakes to Avoid
Wrong: “Line makes $60°$ angle, so slope = $60$”
Correct: Slope $m = \tan 60° = \sqrt{3}$
Remember: Slope is the tangent of the angle, not the angle itself!
Wrong: $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$ (This means coincident lines!)
Correct: $\frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2}$ (For parallel lines)
Key Difference: Equal ratios throughout = same line, equal ratios for A,B only = parallel lines
Common Error: Forgetting the absolute value in distance formula
$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$Remember: Distance is always positive! The absolute value ensures this.
Wrong: Using $\frac{x}{a} + \frac{y}{b} = 1$ when line passes through origin
Correct: If line passes through origin, use $y = mx$ or $Ax + By = 0$
Why? Both intercepts are zero, and we can’t divide by zero!
Family of Lines
Lines Through Intersection
Lines passing through the intersection of $L_1: A_1x + B_1y + C_1 = 0$ and $L_2: A_2x + B_2y + C_2 = 0$:
$$\boxed{L_1 + \lambda L_2 = 0}$$or
$$\boxed{A_1x + B_1y + C_1 + \lambda(A_2x + B_2y + C_2) = 0}$$In simple terms: “Any combination of two intersecting lines gives a line through their intersection point”
JEE Trick: This eliminates finding the intersection point explicitly!
Find the line through intersection of $x + y = 1$ and $2x - y = 5$ that passes through origin.
Solution: Family: $(x + y - 1) + \lambda(2x - y - 5) = 0$
For origin $(0,0)$: $-1 + \lambda(-5) = 0$ → $\lambda = -\frac{1}{5}$
Line: $(x + y - 1) - \frac{1}{5}(2x - y - 5) = 0$
Simplifying: $3x + 6y = 0$ or $x + 2y = 0$
Practice Problems
Level 1: Foundation (NCERT Style)
Question: Find the equation of line passing through $(2, 3)$ and $(4, 7)$.
Solution:
Using two-point form:
$$\frac{y - 3}{7 - 3} = \frac{x - 2}{4 - 2}$$ $$\frac{y - 3}{4} = \frac{x - 2}{2}$$ $$2(y - 3) = 4(x - 2)$$ $$2y - 6 = 4x - 8$$ $$\boxed{4x - 2y - 2 = 0 \quad \text{or} \quad 2x - y - 1 = 0}$$Question: Find equation of line parallel to $3x + 4y = 7$ passing through $(1, 2)$.
Solution:
Parallel lines have same slope.
From $3x + 4y = 7$: slope $m = -\frac{3}{4}$
Using point-slope form:
$$y - 2 = -\frac{3}{4}(x - 1)$$ $$4y - 8 = -3x + 3$$ $$\boxed{3x + 4y = 11}$$Question: Find distance from point $(3, 4)$ to line $5x - 12y + 10 = 0$.
Solution:
Using distance formula:
$$d = \frac{|5(3) - 12(4) + 10|}{\sqrt{5^2 + 12^2}}$$ $$d = \frac{|15 - 48 + 10|}{\sqrt{25 + 144}} = \frac{|-23|}{13} = \frac{23}{13}$$ $$\boxed{d = \frac{23}{13} \text{ units}}$$Level 2: JEE Main Type
Question: Find the acute angle between lines $x + 2y = 3$ and $2x - y = 5$.
Solution:
Slopes: $m_1 = -\frac{1}{2}$, $m_2 = 2$
$$\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| = \left|\frac{-\frac{1}{2} - 2}{1 + (-\frac{1}{2})(2)}\right|$$ $$= \left|\frac{-\frac{5}{2}}{1 - 1}\right| = \left|\frac{-\frac{5}{2}}{0}\right|$$Since denominator is zero, $\tan \theta = \infty$
$$\boxed{\theta = 90°}$$The lines are perpendicular! (Can verify: $m_1 \cdot m_2 = -1$)
Question: Find the reflection of point $(2, 3)$ in the line $x + y = 0$.
Solution:
Let reflection be $(h, k)$.
Condition 1: Midpoint of $(2,3)$ and $(h,k)$ lies on the line:
$$\frac{2+h}{2} + \frac{3+k}{2} = 0$$ $$h + k = -5 \quad \text{...(1)}$$Condition 2: Line joining $(2,3)$ and $(h,k)$ is perpendicular to $x+y=0$ (slope $= -1$):
$$\frac{k-3}{h-2} \times (-1) = -1$$ $$\frac{k-3}{h-2} = 1$$ $$k - 3 = h - 2$$ $$h - k = 1 \quad \text{...(2)}$$From (1) and (2): $h = -2$, $k = -3$
$$\boxed{\text{Reflection: } (-2, -3)}$$Question: Find foot of perpendicular from $(1, 2)$ to line $2x + y = 5$.
Solution:
Perpendicular to $2x + y = 5$ (slope $= -2$) has slope $= \frac{1}{2}$
Perpendicular line through $(1,2)$:
$$y - 2 = \frac{1}{2}(x - 1)$$ $$2y - 4 = x - 1$$ $$x - 2y + 3 = 0 \quad \text{...(1)}$$Foot is intersection of (1) and $2x + y = 5$:
From $2x + y = 5$: $y = 5 - 2x$
Substitute in (1): $x - 2(5-2x) + 3 = 0$
$$x - 10 + 4x + 3 = 0$$ $$5x = 7$$→ $x = \frac{7}{5}$
$$y = 5 - 2 \times \frac{7}{5} = \frac{11}{5}$$ $$\boxed{\text{Foot: } \left(\frac{7}{5}, \frac{11}{5}\right)}$$Level 3: JEE Advanced Type
Question: A line passes through the point $(2, 3)$ and makes intercepts on axes whose sum is zero. Find its equation.
Solution:
Let intercepts be $a$ and $b$ where $a + b = 0$ → $b = -a$
Intercept form: $\frac{x}{a} + \frac{y}{b} = 1$
$$\frac{x}{a} + \frac{y}{-a} = 1$$ $$\frac{x - y}{a} = 1$$ $$x - y = a \quad \text{...(1)}$$Line passes through $(2, 3)$:
$$2 - 3 = a$$ $$a = -1$$From (1): $x - y = -1$
$$\boxed{x - y + 1 = 0}$$Alternate case: If the line passes through origin (both intercepts zero, sum = 0):
Slope through $(2,3)$ and $(0,0)$: $m = \frac{3}{2}$
$$\boxed{y = \frac{3}{2}x \quad \text{or} \quad 3x - 2y = 0}$$Answer: $x - y + 1 = 0$ or $3x - 2y = 0$
Question: Find area of triangle formed by line $3x + 4y = 12$ with coordinate axes.
Solution:
Find intercepts:
- x-intercept (put $y=0$): $3x = 12$ → $x = 4$
- y-intercept (put $x=0$): $4y = 12$ → $y = 3$
Triangle has vertices $(0, 0)$, $(4, 0)$, $(0, 3)$
$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 3$$ $$\boxed{\text{Area} = 6 \text{ sq. units}}$$Formula method: For line $\frac{x}{a} + \frac{y}{b} = 1$, area $= \frac{1}{2}|ab|$
Here: $a = 4$, $b = 3$ → Area $= \frac{1}{2} \times 4 \times 3 = 6$ ✓
Question: Find value of $k$ for which lines $2x + 3y = 5$, $3x + 4y = 6$, and $4x + 5y = k$ are concurrent.
Solution:
Lines are concurrent if they pass through same point.
Find intersection of first two lines:
From $2x + 3y = 5$: $2x = 5 - 3y$ → $x = \frac{5-3y}{2}$
Substitute in $3x + 4y = 6$:
$$3 \times \frac{5-3y}{2} + 4y = 6$$ $$\frac{15 - 9y + 8y}{2} = 6$$ $$15 - y = 12$$ $$y = 3$$ $$x = \frac{5 - 9}{2} = -2$$Intersection point: $(-2, 3)$
For third line to pass through this point:
$$4(-2) + 5(3) = k$$ $$-8 + 15 = k$$ $$\boxed{k = 7}$$Quick Revision Box
| Situation | Formula/Approach |
|---|---|
| Slope from angle | $m = \tan \theta$ |
| Parallel lines | $m_1 = m_2$ or $A_1/A_2 = B_1/B_2 \neq C_1/C_2$ |
| Perpendicular lines | $m_1 m_2 = -1$ or $A_1A_2 + B_1B_2 = 0$ |
| Angle between lines | $\tan\theta = \left |
| Distance: point to line | $d = \frac{ |
| Distance: parallel lines | $d = \frac{ |
| Triangle area | For $\frac{x}{a}+\frac{y}{b}=1$: Area $= \frac{1}{2} |
| Family of lines | $L_1 + \lambda L_2 = 0$ through intersection |
Cross-Links to Related Topics
Prerequisites:
- Trigonometric Identities - For slope and angle conversions
- Inverse Trigonometry - For angle calculations
Related Topics:
- Circles - Lines as tangents to circles
- Parabola - Tangent and normal lines to parabola
- Ellipse - Lines as tangents/chords to ellipse
- Hyperbola - Asymptotes and tangents
- Tangent and Normal to Conics - Unified approach to tangent lines
- Line in Space - Extension to 3D lines
- Vector Basics - Vector form of line equations
Applications:
- Differentiation - Tangent Lines - Finding tangent lines using calculus
- Physics: Kinematics - Motion in a straight line
- Physics: Friction - Inclined planes
JEE Exam Strategy
Weightage: 3-4 questions in JEE Main, 1-2 in Advanced
Most Frequently Asked:
- Distance from point to line (appears almost every year)
- Angle between two lines
- Family of lines through intersection
- Reflection and foot of perpendicular
Time-Saving Tricks:
- Always convert to $Ax + By + C = 0$ for calculations
- For perpendicularity, just check $A_1A_2 + B_1B_2 = 0$ (faster than slopes)
- For parallel distance, ensure same $(A, B)$ coefficients first
Common Trap Options:
- Giving slope value instead of tan of angle
- Missing absolute value in distance formula
- Confusing parallel and coincident line conditions
Teacher’s Summary
- Master the distance formula $d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$ - this appears in 70% of coordinate geometry problems
- Remember slope relationships: Parallel means equal slopes, perpendicular means product = -1
- Use the right form: Choose line equation form based on given information to save time
- Family of lines trick $L_1 + \lambda L_2 = 0$ eliminates tedious intersection calculations
“A straight line is the foundation. Master this, and all curves become easy - because tangents, normals, and asymptotes are all straight lines!”
Next Step: Move to Circles where straight lines become tangents and normals!