Coordinate Geometry bridges algebra and geometry — every point is a pair of numbers, every line is an equation. For NDA, this is one of the most directly testable chapters: distance between points, section formula, equations of lines in five standard forms, angle between lines, and perpendicular distance. Questions are formula-driven and reward students who can identify the correct form quickly from the given information.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Distance between two points and midpoint calculation; (2) Section formula (internal division) for a point dividing a segment in ratio m:n; (3) Area of triangle with given vertices using the determinant formula; (4) Writing equation of a line in slope-intercept, point-slope, or intercept form; (5) Finding slope from two points, or from the equation Ax+By+C=0; (6) Condition for two lines to be parallel (m₁=m₂) or perpendicular (m₁m₂=−1); (7) Distance of a point from a given line using the perpendicular-distance formula; (8) Angle between two lines and family of lines through intersection.
Topics at a Glance
① Coordinate System
Distance, midpoint, section formula, area of triangle
② Slope of a Line
m = (y₂−y₁)/(x₂−x₁) = tanθ
③ Five Forms of Line
Slope-intercept, point-slope, two-point, intercept, general
④ Parallel & Perpendicular
m₁=m₂ (parallel); m₁m₂=−1 (perp)
⑤ Angle & Distance
Angle between lines; distance from point to line
⑥ Family of Lines
L₁+λL₂=0 through intersection
1. Coordinate System — Distance, Section & Area
1.1
Distance Formula, Section Formula & Midpoint
The three foundational results — everything else builds on these
⚡ Distance, Midpoint & Section Formulas
DISTANCE between P(x₁,y₁) and Q(x₂,y₂):
PQ = √[(x₂−x₁)² + (y₂−y₁)²]
MIDPOINT of PQ:
M = ((x₁+x₂)/2, (y₁+y₂)/2)
SECTION FORMULA — point dividing PQ in ratio m:n:
Internal division:
R = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n))
External division:
R = ((mx₂−nx₁)/(m−n), (my₂−ny₁)/(m−n))
Centroid of triangle with vertices (x₁,y₁),(x₂,y₂),(x₃,y₃):
G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)
Section formula memory: “m times far point plus n times near point, over (m+n)”. Internal means the point lies BETWEEN P and Q. External means it lies OUTSIDE on the extension.
⚡ Area of a Triangle Using Coordinates
Area of triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃):
Δ = ½ |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|
Equivalent determinant form:
Δ = ½ |x₁ y₁ 1|
|x₂ y₂ 1|
|x₃ y₃ 1|
Three points are COLLINEAR when Δ = 0:
x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂) = 0
The absolute value is essential — area is always positive. The collinearity condition is a direct NDA MCQ: given three points, check if the area = 0. The formula works for any orientation of the triangle.
Fig 1: Points P, Q on coordinate plane. M=midpoint (amber), R=internal division point (purple). Right: triangle ABC with area formula.
Worked Example — Section Formula
Find the point that divides the segment joining A(2,−1) and B(8,5) internally in ratio 2:1.
x = (2×8+1×2)/(2+1) = (16+2)/3 = 18/3 = 6.
y = (2×5+1×(−1))/(2+1) = (10−1)/3 = 9/3 = 3.
Point = (6, 3).
Worked Example — Area of Triangle
Find the area of the triangle with vertices A(1,2), B(4,6), C(7,2).
Δ = ½ |1(6−2)+4(2−2)+7(2−6)| = ½ |4+0−28| = ½ × 24 = 12 sq. units.
📝 TOPIC-WISE PYQ
Coordinate Basics — NDA-Pattern Questions
Q1. The distance between points (3, 4) and (7, 1) is:
Slope = tanθ where θ is the angle the line makes with the positive x-axis
⚡ Slope — All Forms
Slope from two points (x₁,y₁) and (x₂,y₂):
m = (y₂−y₁) / (x₂−x₁) (x₁ ≠ x₂)
Slope from inclination angle θ:
m = tan θ (where 0 ≤ θ < 180°, θ ≠ 90°)
Slope from general equation Ax+By+C = 0:
m = −A/B (provided B ≠ 0)
y-intercept c = −C/B (rearrange to y=mx+c form)
Special cases:
Horizontal line (y = k): m = 0 (tan0° = 0)
Vertical line (x = k): m is undefined (tan90° = ∞)
Line through origin: c = 0, so y = mx
Line making 45° with x-axis: m = 1
Line making 135° with x-axis: m = −1
From general form Ax+By+C=0: slope = −A/B. This is the fastest way to find slope without rearranging. Always divide through by B to reach y=mx+c form when needed.
Slope Values for Standard Angles
0°: m = 0 (horizontal)
30°: m = 1/√3 ≈ 0.577
45°: m = 1
60°: m = √3 ≈ 1.732
90°: m undefined (vertical)
120°: m = −√3 (tan120°)
135°: m = −1
150°: m = −1/√3
Slope Sign Interpretation
m > 0: line rises left to right (acute angle)
m < 0: line falls left to right (obtuse angle)
m = 0: horizontal line
|m| large: steep line (near vertical)
|m| small: gentle slope (near horizontal)
Same slope ⇒ parallel lines
Product −1 ⇒ perpendicular lines
3. Five Standard Forms of Equation of a Line
3.1
All Five Forms — When to Use Each
Choose the form that matches what you are given in the problem
① Slope–Intercept
y = mx + c
Use when: slope m and y-intercept c are given. Fastest to sketch. m = slope, c = where line cuts y-axis.
② Point–Slope
y − y₁ = m(x − x₁)
Use when: slope m and one point (x₁,y₁) are known. Most versatile in NDA problems.
③ Two–Point Form
(y−y₁)/(y₂−y₁) = (x−x₁)/(x₂−x₁)
Use when: two points (x₁,y₁) and (x₂,y₂) are given. Find slope first, then use point-slope.
④ Intercept Form
x/a + y/b = 1
Use when: x-intercept a and y-intercept b are given. Line cuts x-axis at (a,0) and y-axis at (0,b).
⑤ General Form
Ax + By + C = 0
The universal form. slope=−A/B, x-int=−C/A, y-int=−C/B. Used for distance and angle formulas.
📌 Quick Conversion
Ax+By+C=0 ⇔ y = (−A/B)x + (−C/B)
Rearranging to slope-intercept gives m=−A/B, c=−C/B. Always check sign carefully.
Fig 2: Left — Slope-intercept form y=mx+c: inclination angle θ (m=tanθ) and y-intercept labelled c. Right — Intercept form x/a+y/b=1: x-intercept a and y-intercept b are distinct labels for distinct forms.
Worked Example — Writing Line in Multiple Forms
Find the equation of the line through (2, 3) with slope −2.
Q6. The equation of the line through (1, 2) and (3, 6) is:
(a) y = 2x (b) y = x+1 (c) 2x−y=0 (d) x+y=3
Answer: (a) y = 2x
m=(6−2)/(3−1)=4/2=2. Using point (1,2): y−2=2(x−1) → y=2x. Check: (0,0) on line means it passes through origin → y=2x.
Q7. The y-intercept of the line 5x + 2y − 10 = 0 is:
(a) 2 (b) 5 (c) −5 (d) −2
Answer: (b) 5
c = −C/B = −(−10)/2 = 5. Or set x=0: 2y=10 → y=5.
4. Parallel & Perpendicular Lines, Angle Between Lines
4.1
Conditions for Parallel & Perpendicular, and Angle Formula
Three of the most directly tested results in NDA straight-line questions
⚡ Parallel, Perpendicular & Angle Between Lines
Two lines with slopes m₁ and m₂:
PARALLEL (same direction, never meet):
m₁ = m₂ (slopes equal, intercepts different)
In general form: A₁/A₂ = B₁/B₂ ≠ C₁/C₂
PERPENDICULAR (meet at 90°):
m₁ × m₂ = −1 (product of slopes = −1)
OR: m₂ = −1/m₁ (negative reciprocal)
In general form: A₁A₂ + B₁B₂ = 0
ANGLE between lines with slopes m₁, m₂:
tan θ = |(m₁−m₂) / (1+m₁m₂)|
If 1+m₁m₂ = 0: lines are perpendicular (θ=90°)
If m₁ = m₂: lines are parallel (θ=0°)
Perpendicular line to Ax+By+C=0 through (x₁,y₁):
B(x−x₁) − A(y−y₁) = 0 (swap A and B, change one sign)
Parallel line to Ax+By+C=0:
Ax + By + K = 0 (same A and B, different constant K)
Angle formula is identical to the compound-angle tan(A−B) formula from trigonometry. For NDA, the most common use is checking parallel (m₁=m₂) or perpendicular (m₁m₂=−1) conditions.
Parallel Line Pairs
2x+3y=5 and 2x+3y=7: parallel (same A,B)
y=3x+1 and y=3x−4: parallel (same m=3)
Parallel through (a,b): Ax+By=Aa+Bb
Check: ratio A/A=B/B, but C₁≠C₂
Distance between parallel lines: |C₁−C₂|/√(A²+B²)
Perpendicular Line Pairs
y=2x+3 and y=−x/2+1: perp (2×−1/2=−1 ✓)
Perp to 3x+4y=7 through origin: 4x−3y=0
Perp to y=mx: slope = −1/m
Horizontal (m=0) ⊥ vertical (m=undefined)
A₁A₂+B₁B₂=0 (general form check)
Worked Example — Angle Between Lines
Find the acute angle between y=2x+1 and y=3x+2.
m₁=2, m₂=3. tanθ = |(2−3)/(1+6)| = |−1/7| = 1/7.
θ = tan⁻¹(1/7) ≈ 8.1°.
5. Distance of a Point from a Line
5.1
Perpendicular Distance Formula & Distance Between Parallel Lines
Always convert to general form Ax+By+C=0 before applying the formula
⚡ Perpendicular Distance Formulas
Distance from point P(x₀,y₀) to line Ax+By+C=0:
d = |Ax₀ + By₀ + C| / √(A²+B²)
Note: The numerator uses absolute value → distance is always positive.
The denominator √(A²+B²) is the “norm” of the line.
Distance from ORIGIN (0,0) to Ax+By+C=0:
d = |C| / √(A²+B²)
Distance between two PARALLEL lines Ax+By+C₁=0 and Ax+By+C₂=0:
d = |C₁−C₂| / √(A²+B²)
FOOT of perpendicular from P(x₀,y₀) to Ax+By+C=0:
(h,k) where: (h−x₀)/A = (k−y₀)/B = −(Ax₀+By₀+C)/(A²+B²)
The formula always needs the line in Ax+By+C=0 form. If given y=mx+c, rewrite as mx−y+c=0 (so A=m, B=−1, C=c) and then apply the formula.
Worked Example — Distance from Point to Line
Find the distance from (3, −4) to the line 3x − 4y + 5 = 0.
A=3, B=−4, C=5, x₀=3, y₀=−4.
d = |3(3)+(−4)(−4)+5|/√(9+16) = |9+16+5|/√25 = 30/5 = 6.
Worked Example — Distance Between Parallel Lines
Find the distance between 3x+4y−5=0 and 3x+4y+15=0.
🤯 T1. Find the equation of the line passing through the intersection of 2x+y=5 and x−y=1, and perpendicular to x−2y+3=0.
Intersection: add the equations: 3x=6 → x=2. y=5−4=1. Point=(2,1).
Line x−2y+3=0 has slope m₁=1/2.
Perpendicular slope: m₂=−1/m₁=−2.
Line through (2,1) with slope −2: y−1=−2(x−2) → y=−2x+5.
General form: 2x+y−5=0. Two-step: (1) find intersection point, (2) write line through that point with the required slope.
🤯 T2. Show that the three lines 2x+y=5, x−y=1, and x+y=5 form a right triangle and find the area.
🤯 T3. Find the value of k so that 2x+3y=7 and 2kx+(k+1)y=28 have no solution (i.e., are parallel).
Parallel condition: A₁/A₂ = B₁/B₂ ≠ C₁/C₂.
2/(2k) = 3/(k+1) ⇒ 2(k+1) = 3(2k) ⇒ 2k+2 = 6k ⇒ 4k=2 ⇒ k=1/2.
Verify C ratio: 7/28 = 1/4 ≠ 1/2 = A₁/A₂ ✓ (not the same, so lines are indeed parallel, no solution). k = 1/2. For no solution: same slope (parallel), different intercepts. For infinite solutions: all three ratios equal (coincident lines).
6. Family of Lines Through Intersection
6.1
Equation L₁ + λL₂ = 0 — Family of Lines Concept
Any line through the intersection of L₁=0 and L₂=0 can be written as L₁+λL₂=0
⚡ Family of Lines — Through Intersection
If L₁: A₁x+B₁y+C₁ = 0 and L₂: A₂x+B₂y+C₂ = 0 intersect at point P:
Family of ALL lines through P:
L₁ + λL₂ = 0 for any real value of λ
⇔ (A₁+λA₂)x + (B₁+λB₂)y + (C₁+λC₂) = 0
USE:
To find a specific member of the family, apply an additional condition
(e.g., passes through another given point, or has a given slope),
solve for λ, then substitute back.
KEY OBSERVATIONS:
λ=0 gives L₁ itself.
λ→∞ gives L₂ itself.
Every finite value of λ gives a new line through the intersection.
This is a powerful shortcut. Instead of finding the intersection point and then using point-slope, directly write L₁+λL₂=0 and apply the additional condition to find λ. Saves significant time in NDA.
Worked Example — Family of Lines
Find the line through the intersection of x+2y−3=0 and 2x−y+1=0 that passes through (2,0).