Revision of Important Coordinate Geometry Formulas
Before starting Straight Lines, it is essential to revise fundamental coordinate geometry concepts. These formulas form the computational backbone for solving problems related to slope, intercepts, equations of lines, and geometric interpretation in both board and competitive examinations.
Exam Significance
- Direct application in JEE Main and NEET numerical problems
- Used repeatedly in derivation-based board questions
- Foundation for advanced topics like Circles and Conic Sections
- Frequently used in coordinate proofs and geometry transformations
-
Distance Between Two Points
Definition: Length of the line segment joining two points in Cartesian plane.\[ \mathrm{PQ}=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \]Key Insight: Derived using Pythagoras theorem.Example: Find distance between (1,2) and (4,6)
Solution: √[(4−1)² + (6−2)²] = √(9 + 16) = 5 -
Section Formula (Internal Division)
Used to find a point dividing a line segment in a given ratio.\[ \left(\dfrac{mx_2+nx_1}{m+n},\; \dfrac{my_2+ny_1}{m+n}\right) \]Concept: Weighted average of coordinates. -
Midpoint Formula
Special case of section formula when m = n.\[ \left(\dfrac{x_2+x_1}{2},\; \dfrac{y_2+y_1}{2}\right) \]Use Case: Finding centre of line segment, symmetry problems. -
Area of Triangle (Coordinate Geometry)
\[ \dfrac{1}{2}\Bigl|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\Bigr| \]Important Result: If area = 0 → points are collinear.Example: Check collinearity of (1,2), (2,4), (3,6)
Area = 0 → Points are collinear
Concept Booster for Competitive Exams
- Distance formula is frequently used in locus problems
- Section formula appears in vector and 3D geometry later
- Area formula is a quick test for collinearity in MCQs
- Midpoint is used in symmetry and reflection-based questions
Slope of a Line (Gradient)
The slope of a line quantifies its steepness and direction. If a line makes an angle θ with the positive direction of the x-axis, then the slope (or gradient) is defined as the tangent of that angle.
- Positive slope → line rises from left to right
- Negative slope → line falls from left to right
- Zero slope → horizontal line (parallel to x-axis)
- Undefined slope → vertical line (parallel to y-axis)
| Line Type | Slope |
|---|---|
| x-axis | 0 |
| y-axis | Not defined |
- Slope determines parallelism: equal slopes → parallel lines
- Product of slopes = -1 → perpendicular lines
- Used in derivative interpretation in calculus (rate of change)
What is the slope of a vertical line?
Slope of a Line Using Coordinates of Two Points
The slope of a line passing through two distinct points represents the rate of change of y with respect to x. It is one of the most fundamental concepts in coordinate geometry and is extensively used in straight lines, calculus, and physics.
- Numerator (y₂ − y₁) → vertical change (rise)
- Denominator (x₂ − x₁) → horizontal change (run)
- Slope = change in y per unit change in x
- If y₂ > y₁ → slope positive
- If y₂ < y₁ → slope negative
- If x₂ = x₁ → slope undefined (vertical line)
- If y₂ = y₁ → slope zero (horizontal line)
Find slope between (1,2) and (3,6)
- Slope is invariant under translation of coordinate system
- Used to determine angle between two lines
- Foundation for equation forms: point-slope, slope-intercept
- Critical in JEE problems involving locus and transformations
Conditions for Parallelism and Perpendicularity Using Slopes
The relationship between two straight lines in a plane can be completely determined using their slopes. This concept is fundamental in coordinate geometry and plays a crucial role in solving problems related to angles, intersections, and geometric configurations in board exams as well as competitive exams like JEE and NEET.
Condition for Parallel Lines
Two lines are parallel if they have the same inclination and therefore never intersect. This happens when their slopes are equal.
Condition for Perpendicular Lines
Two lines are perpendicular if they intersect at a right angle (90°). This occurs when the product of their slopes is equal to -1.
Important Edge Cases
- Vertical line ⟂ Horizontal line (always perpendicular)
- Two vertical lines are parallel
- Two horizontal lines are parallel
- Vertical line slope is undefined, so product rule cannot be directly applied
Are lines with slopes 4 and -1/4 perpendicular?
Competitive Exam Insights
- Used to prove perpendicularity in coordinate proofs
- Frequently appears in JEE problems involving triangles and circles
- Essential for angle between two lines formula
- Helps identify orthogonality in vectors and 3D geometry
If m₁ = m₂ → Lines are parallel
If m₁ × m₂ = -1 → Lines are perpendicular
Angle Between Two Lines
The angle between two straight lines is the smallest angle through which one line must be rotated about their point of intersection to coincide with the other. This concept is fundamental in coordinate geometry and widely used in solving problems involving slopes, perpendicularity, and geometric configurations.
- Each line makes an angle (inclination) with x-axis
- Slope = tan(inclination)
- Angle between lines = difference of inclinations
- Apply tan(A − B) identity to derive formula
- If m₁ = m₂ → θ = 0° (parallel lines)
- If m₁m₂ = -1 → θ = 90° (perpendicular lines)
- If denominator = 0 → angle is 90°
- Used in JEE problems involving triangle geometry in coordinate plane
- Helps verify perpendicularity quickly
- Frequently appears in multi-step coordinate geometry problems
- Foundation for angle between planes and vectors (Class 12)
tanθ = |(m₂ − m₁)/(1 + m₁m₂)|
Parallel → θ = 0°
Perpendicular → θ = 90°
Various Forms of the Equation of a Line
A straight line can be represented algebraically in different forms depending on the given information such as slope, intercepts, or points. Understanding all forms is essential for solving coordinate geometry problems efficiently in board examinations and competitive exams like JEE and NEET.
Horizontal and Vertical Lines
A horizontal line is parallel to x-axis and vertical line is parallel to y-axis.
\[ y = y_1 \quad , \quad x = x_1 \]- Horizontal line → slope = 0
- Vertical line → slope undefined
Point–Slope Form
Used when slope and one point are known.
\[ y - y_1 = m(x - x_1) \]⇒ y − 3 = 2(x − 1)
Two–Point Form
Used when two points are given.
\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]Slope–Intercept Form
Represents a line using slope and y-intercept.
\[ y = mx + c \]- m → slope
- c → y-intercept
Intercept Form
Represents a line using x and y intercepts.
\[ \frac{x}{a} + \frac{y}{b} = 1 \]Conversion Between Forms
- Point-slope → slope-intercept by simplification
- Two-point → find slope then apply point-slope
- Intercept → convert to slope-intercept form
Exam Insights
- JEE: Switching forms quickly saves time
- Boards: Derivations often asked
- Graphs: slope-intercept form most useful
- Geometry: intercept form gives visual clarity
All forms represent the same straight line. Choosing the right form simplifies calculations and improves problem-solving efficiency.
Distance of a Point From a Line
The distance of a point from a line is defined as the length of the perpendicular drawn from the point to the line. This represents the shortest possible distance between the point and the line in a coordinate plane.
- Expression (ax₁ + by₁ + c) → measures deviation from line
- Denominator normalizes direction using slope
- Absolute value ensures distance is always positive
The formula is derived using projection of the vector joining the point and the line onto the normal vector (a, b) of the line. Hence, it represents perpendicular distance.
- If ax₁ + by₁ + c = 0 → point lies on the line → distance = 0
- For horizontal line (y = k) → distance = |y₁ − k|
- For vertical line (x = k) → distance = |x₁ − k|
d = |3(1) + 4(2) + 5| / √(9 + 16)
= |3 + 8 + 5| / 5 = 16/5
Applications in Exams
- Finding shortest distance in coordinate geometry
- Checking if a point lies on a line
- Used in locus problems
- Applied in optimization problems in JEE
Distance is always measured perpendicular to the line and is given by a normalized algebraic expression.
Distance Between Two Parallel Lines
The distance between two parallel lines is the length of the perpendicular drawn from any point on one line to the other. Since parallel lines never intersect, this distance remains constant throughout.
- Distance is independent of chosen point
- Only constant terms (c₁, c₂) affect separation
- Denominator normalizes direction of line
If lines are already in same normalized form (same a, b), directly apply formula without finding any point.
d = |4 - (-5)| / √(4 + 9) = 9 / √13
- If c₁ = c₂ → lines coincide → distance = 0
- If coefficients differ → first convert to same form
- Applicable only for parallel lines
Competitive Exam Insights
- Frequently asked in JEE as direct formula-based question
- Used in optimization and locus problems
- Helps in finding distance between boundaries in geometry
- Can be combined with area and triangle problems
Distance between parallel lines depends only on difference of constants and remains uniform everywhere.
Example 1: Finding Slope of Lines
Find the slope of the lines:
(a) Through (3, –2) and (–1, 4)
(b) Through (3, –2) and (7, –2)
(c) Through (3, –2) and (3, 4)
(d) Inclination 60°
\[ m = \frac{y_2 - y_1}{x_2 - x_1}, \quad m = \tan\theta \]
(a) General Case (Oblique Line)
Conclusion: Negative slope → line decreases from left to right
(b) Horizontal Line
Conclusion: Slope = 0 → line parallel to x-axis
(c) Vertical Line
Conclusion: Undefined slope → vertical line
(d) Using Inclination
Conclusion: Positive steep slope
What type of line has undefined slope?
Exam Insights
- Check denominator first → avoid division errors
- Equal y → slope = 0 (horizontal)
- Equal x → slope undefined (vertical)
- Inclination problems directly use tanθ
(a) m = -3/2
(b) m = 0
(c) Undefined
(d) m = √3
Example 2: Finding Slope Using Angle Between Lines
If the angle between two lines is π/4 and slope of one line is 1/2, find the slope of the other line.
\[ \tan\theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right| \]
(i) Case: Expression = +1
(ii) Case: Expression = −1
- Absolute value ⇒ two possible solutions
- Both lines form same angle but different orientations
- Both answers valid unless restriction given
If angle = 90°, what must be true?
m = 3 or m = −1/3
Exam Insights
- Always consider ± cases (very important)
- Common JEE pattern question
- Use shortcut: perpendicular ⇒ m₁m₂ = −1
Example 3: Perpendicular Lines & Finding Unknown Coordinate
Line through (–2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (x, 24). Find x.
For perpendicular lines: \[ m_1 \cdot m_2 = -1 \]
Step 1: Find slope of first line
Step 2: Find slope of second line
Step 3: Apply perpendicular condition
- Always compute slopes separately before applying condition
- Keep unknown variable symbolic until final step
- Perpendicular condition simplifies equation quickly
If slope of one line is 2, what must be slope of perpendicular line?
x = 4
Exam Insights
- Very common pattern: unknown coordinate + perpendicular condition
- Always simplify fractions early to avoid mistakes
- Watch sign carefully when solving equations
- Shortcut: m₂ = -1/m₁ (if easier)
Example 4: Lines Parallel to Coordinate Axes
Find the equations of the lines parallel to axes and passing through (–2, 3).
Line ∥ x-axis → y = constant
Line ∥ y-axis → x = constant
Line Parallel to y-axis
A line parallel to the y-axis has a fixed x-coordinate. Since the point is (–2, 3), the x-value remains constant.
Line Parallel to x-axis
A line parallel to the x-axis has a fixed y-coordinate. Since the point is (–2, 3), the y-value remains constant.
- Parallel to x-axis → slope = 0
- Parallel to y-axis → slope undefined
- Such equations are the simplest form of straight lines
Which equation represents a vertical line?
x = -2 and y = 3
Exam Insights
- Very common basic concept question in boards
- Used as base case in coordinate geometry problems
- Important for graph plotting and visualization
- Often used in combination with distance formulas
Example 5: Equation of Line Using Point–Slope Form
Find the equation of the line through (–2, 3) with slope –4.
\[ y - y_1 = m(x - x_1) \]
Step 1: Substitute values
Step 2: Simplify
You can directly use slope-intercept form y = mx + c. Substitute point (–2,3) to find c:
Substitute point (–2,3) into y = –4x – 5:
If slope is positive, the line will:
y = -4x - 5
Exam Insights
- Point-slope form is fastest when slope & one point are given
- Always simplify to y = mx + c (preferred final form)
- Verification step avoids sign mistakes
- Very common in boards and JEE
Example 6: Equation of Line Through Two Points
Write the equation of the line through the points (1, –1) and (3, 5).
\[ m = \frac{y_2 - y_1}{x_2 - x_1}, \quad y - y_1 = m(x - x_1) \]
Step 1: Find slope
Step 2: Use point–slope form
Check both points:
- Two distinct points always determine a unique line
- Either point can be used in point-slope form
- Final answer should be simplified to y = mx + c
If slope = 0, the line is:
y = 3x - 4
Exam Insights
- Two-point problems are very common in boards and JEE
- Always calculate slope carefully to avoid sign errors
- Direct formula saves time in MCQs
- Verification ensures correctness
Example 7: Equation Using Inclination and Intercepts
Write the equation of the lines for which tanθ = 1/2, where θ is inclination, and:
(i) y-intercept = -3/2
(ii) x-intercept = 4
(i) Line with y-intercept = -3/2
\[ y = \frac{1}{2}x - \frac{3}{2} \] \[ 2y - x + 3 = 0 \](ii) Line with x-intercept = 4
\[ y = \frac{1}{2}(x - 4) \] \[ y = \frac{1}{2}x - 2 \] \[ 2y - x + 4 = 0 \]- Both lines have same slope → parallel lines
- Intercept changes vertical position
- Choose equation form based on given data
(i) 2y - x + 3 = 0
(ii) 2y - x + 4 = 0
Example 8: Equation Using Intercept Form
Find the equation of the line which makes intercepts –3 and 2 on the x- and y-axes respectively.
\[ \frac{x}{a} + \frac{y}{b} = 1 \]
Step 1: Substitute intercepts
Step 2: Simplify
- Intercepts directly define where line meets axes
- Sign of intercept is very important (left/right, up/down)
- Intercept form is fastest when both intercepts are given
If both intercepts are positive, the line lies in:
3y - 2x - 6 = 0
Exam Insights
- Intercept form is commonly asked in boards
- Watch signs carefully (very common mistake)
- Always convert to standard form for final answer
- Useful in graph-based MCQs
Example 9: Distance of a Point from a Line
Find the distance of the point (3, –5) from the line 3x – 4y – 26 = 0.
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Step 1: Identify values
Step 2: Substitute
- Always use absolute value in numerator
- Distance is always positive
- Denominator normalizes direction of line
If numerator becomes zero → point lies on line → distance = 0
If A² + B² = 25, denominator becomes:
d = 3/5
Exam Insights
- Direct formula-based question → easy scoring
- Sign errors are very common → be careful
- Frequently appears in JEE and boards
- Can be extended to shortest distance problems
Example 10: Distance Between Parallel Lines
Find the distance between the parallel lines 3x – 4y + 7 = 0 and 3x – 4y + 5 = 0.
\[ d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} \]
Step 1: Identify coefficients
Step 2: Apply formula
- Same coefficients of x and y → lines are parallel
- Only constants (c₁, c₂) determine distance
- Distance is uniform everywhere
If coefficients are identical, directly use difference of constants.
If c₁ = c₂ → lines coincide → distance = 0
If numerator becomes zero, distance is:
d = 2/5
Exam Insights
- Very common direct formula-based question
- Fast solving possible using shortcut
- Check coefficients first before applying formula
- Frequently appears in JEE and boards
Example 11: Concurrent Lines & Finding Parameter
If the lines 2x + y − 3 = 0, 5x + ky − 3 = 0 and 3x − y − 2 = 0 are concurrent, find k.
Three lines are concurrent if they intersect at a single point.
Step 1: Find intersection of first two lines
Intersection Point: (1, 1)
Step 2: Substitute into third line
For concurrency: \[ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0 \] Useful in advanced problems.
- Find intersection of any two lines first
- Substitute into third line
- Determinant method is faster for complex cases
If three lines meet at one point, they are called:
k = -2
Exam Insights
- Concurrency problems are common in JEE
- Intersection method is easiest for beginners
- Determinant method is fastest in MCQs
- Check calculations carefully (common sign errors)
Example 12: Distance Along a Given Direction
Find the distance of the line 4x – y = 0 from the point P(4, 1), measured along a line making 135° with the positive x-axis.
Distance is measured along a given direction → find intersection of direction-line with given line → compute distance between points.
Step 1: Find slope of direction line
Step 2: Equation of line through P(4,1)
Step 3: Find intersection with given line
Step 4: Distance between points
- This is NOT perpendicular distance
- Distance depends on given direction
- Always form line through point with given slope
- Then find intersection → then distance
If θ = 90°, distance becomes:
d = 3√2
Exam Insights
- Advanced question type (JEE level)
- Always identify: perpendicular or oblique distance
- Convert direction → slope using tanθ
- Follow: line → intersection → distance
Example 13: Image of a Point in a Line (Reflection)
Find the image of the point (1, 2) in the line x – 3y + 4 = 0.
Reflection means the given line is the perpendicular bisector of the point and its image.
Step 1: Identify values
Step 2: Compute key terms
Step 3: Apply formula
Midpoint of P and P' should lie on the line:
Substitute in line → satisfies equation ✔
- Line is perpendicular bisector of P and image
- Distance from line is same for both points
- Midpoint always lies on mirror line
Reflection preserves:
(6/5, 7/5)
Exam Insights
- High-weightage JEE topic
- Formula-based but concept-heavy
- Midpoint check is powerful verification
- Used in reflection and symmetry problems
Example 14: Area of Triangle Formed by Lines
Show that the area of the triangle formed by the lines y = m₁x + c₁, y = m₂x + c₂ and x = 0 is:
\[
\frac{(c_1 - c_2)^2}{2|m_1 - m_2|}
\]
Triangle is formed between two lines and the y-axis → use base-height approach.
Step 1: Points on y-axis
Base AB lies along y-axis
Step 2: Intersection of two lines
This gives vertex C
Step 3: Base and Height
Step 4: Area
- Base lies on y-axis → depends only on intercepts
- Height depends on slope difference
- If slopes equal → lines parallel → no triangle
Larger difference in slopes → smaller triangle width → area changes inversely.
If m₁ = m₂, area becomes:
Area = (c₁ − c₂)² / (2 |m₁ − m₂|)
Exam Insights
- Standard derivation question in boards
- Frequently used in JEE problems
- Memorize final formula for quick solving
- Check slopes first to ensure triangle exists
Example 15: Line Bisecting a Segment Between Two Lines
A line is such that its segment between the lines 5x – y + 4 = 0 and 3x + 4y – 4 = 0 is bisected at (1, 5). Find its equation.
If a line cuts two given lines and midpoint is known → intersection points satisfy midpoint formula.
Step 1: Assume equation
Step 2: Intersection with given lines
Let intersections be A and B.
Step 3: Apply midpoint condition
Step 4: Final equation
- Assume line using slope form
- Find intersections with given lines
- Apply midpoint condition
- Solve for slope → final equation
In many problems, symmetry or inspection may directly suggest slope.
If midpoint is known, which formula is used?
y = x + 4
Exam Insights
- Advanced JEE problem type
- Requires multi-step reasoning
- Key step: midpoint condition
- Algebra simplification is crucial
Example 16: Locus of a Point Equidistant from Two Lines
Show that the path of a moving point whose distances from the lines 3x – 2y = 5 and 3x + 2y = 5 are equal is a straight line.
Points equidistant from two lines lie on their angle bisectors.
Step 1: Convert to standard form
Step 2: Use distance formula
Step 3: Solve modulus cases
- Equal distance → angle bisectors
- Always leads to two lines (internal & external bisectors)
- These are straight lines → hence locus is linear
Directly use: \[ \frac{A_1x + B_1y + C_1}{\sqrt{A_1^2 + B_1^2}} = \pm \frac{A_2x + B_2y + C_2}{\sqrt{A_2^2 + B_2^2}} \]
Equal distance from two lines gives:
Locus: y = 0 and x = 5/3
Exam Insights
- Very important JEE concept
- Always results in two straight lines
- Recognize pattern → saves time
- Used in locus and geometry problems
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