Coordinate geometry turns geometry into algebra: once a line lives on the Cartesian plane you can find it, slope it, measure it and intersect it using simple formulas. For NDA Maths, Straight Lines is a high-yield chapter — 3 to 5 questions appear every year on slope, line forms, distance and the pair of straight lines. Master a handful of formulas and these marks are yours.
Why Straight Lines Matters for NDA
Almost every NDA Maths paper carries several questions from coordinate geometry, and the straight line is its backbone. The questions are usually short and formula-driven — find the slope, find the distance of a point from a line, find where two lines meet, or check whether a second-degree equation breaks into two lines. Once you know the formula, the answer falls out in seconds.
This chapter also feeds later topics: circles, conics and 3D geometry all rest on the same Cartesian idea of describing shapes with coordinates and equations. So time spent here pays off across the whole syllabus.
A straight line is the graph of a first-degree equation in x and y, written as ax + by + c = 0. Any equation where the highest power of x and y together is 1 is a line — nothing else.
For a future officer, coordinate geometry is the same maths used in mapping, navigation and gunnery — locating a point by two numbers and finding directions and distances between them.
The Cartesian System and Basic Formulas
The Cartesian plane is formed by two number lines crossing at right angles: the horizontal x-axis and the vertical y-axis, meeting at the origin O(0, 0). Any point is named by an ordered pair (x, y), where x is the abscissa and y is the ordinate.
The two axes cut the plane into four quadrants, numbered anticlockwise:
- Quadrant I: x > 0, y > 0 (both positive)
- Quadrant II: x < 0, y > 0
- Quadrant III: x < 0, y < 0 (both negative)
- Quadrant IV: x > 0, y < 0
For two points A(x1, y1) and B(x2, y2):
- Distance: AB = √[(x2 − x1)2 + (y2 − y1)2]
- Midpoint: M = ( (x1 + x2) ÷ 2 , (y1 + y2) ÷ 2 )
- Section formula (dividing AB in ratio m : n internally): ( (mx2 + nx1) ÷ (m + n) , (my2 + ny1) ÷ (m + n) )
The centroid of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is simply the average: G = ( (x1 + x2 + x3) ÷ 3 , (y1 + y2 + y3) ÷ 3 ).
The area of a triangle with these vertices is ½ × |x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)|. If this area is zero, the three points are collinear — they lie on one straight line.
Slope - The Steepness of a Line
The slope (or gradient), written m, measures how steep a line is and in which direction it tilts. If a line makes an angle θ with the positive x-axis, then m = tan θ.
Slope through two points A(x1, y1) and B(x2, y2):
m = (y2 − y1) ÷ (x2 − x1) (rise ÷ run)
What the slope tells you at a glance:
- m > 0 → line goes up from left to right.
- m < 0 → line goes down from left to right.
- m = 0 → horizontal line (parallel to x-axis).
- m undefined → vertical line (parallel to y-axis), since x2 − x1 = 0.
For two lines with slopes m1 and m2:
- Parallel when m1 = m2.
- Perpendicular when m1 × m2 = −1.
The slope of a line given as ax + by + c = 0 is m = −a ÷ b. You do not need to rearrange — just read off the coefficients. This single trick answers many parallel/perpendicular PYQs in one step.
Forms of the Equation of a Line
The same line can be written in several standard forms. Choose the one that matches the data given in the question.
Slope-intercept form
y = mx + c, where m is the slope and c is the y-intercept (where the line cuts the y-axis).
Point-slope form
y − y1 = m(x − x1), used when you know one point (x1, y1) and the slope m.
Two-point form
y − y1 = [(y2 − y1) ÷ (x2 − x1)] (x − x1), used when two points are known.
Intercept form
(x ÷ a) + (y ÷ b) = 1, where a is the x-intercept and b is the y-intercept.
Normal (perpendicular) form
x cos θ + y sin θ = p, where p is the perpendicular distance of the line from the origin and θ is the angle that perpendicular makes with the x-axis.
A horizontal line is y = k; a vertical line is x = k. The general form ax + by + c = 0 can hold every case, including vertical lines that y = mx + c cannot.
In the intercept form, a and b are the actual intercepts, not the coefficients. From 2x + 3y = 6, divide by 6 to get (x ÷ 3) + (y ÷ 2) = 1, so the x-intercept is 3 and the y-intercept is 2 — not 2 and 3.
Distance of a Point and Angle Between Lines
Two formulas in this section appear again and again in the NDA exam, so commit them to memory.
Perpendicular distance of a point (x1, y1) from the line ax + by + c = 0:
d = |ax1 + by1 + c| ÷ √(a2 + b2)
The distance between two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is |c1 − c2| ÷ √(a2 + b2). Make sure the coefficients of x and y are identical in both equations before using it.
Angle θ between two lines of slopes m1 and m2:
tan θ = |(m1 − m2) ÷ (1 + m1m2)|
From this single formula both special cases drop out: if m1 = m2 the numerator is 0 so θ = 0° (parallel), and if 1 + m1m2 = 0 the denominator is 0 so θ = 90° (perpendicular).
In the distance formula, the modulus bars mean distance is always positive. If a question gives a point and asks how far it is from a line, plug straight into d = |ax1 + by1 + c| ÷ √(a2 + b2) — do not try to draw it.
Pair of Straight Lines
A second-degree equation can sometimes represent two straight lines at once. The general second-degree equation is:
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
Lines through the origin
The homogeneous equation ax2 + 2hxy + by2 = 0 always represents a pair of straight lines passing through the origin. If their slopes are m1 and m2, then:
- Sum of slopes: m1 + m2 = −2h ÷ b
- Product of slopes: m1m2 = a ÷ b
The angle θ between the pair of lines ax2 + 2hxy + by2 = 0 is:
tan θ = |2√(h2 − ab) ÷ (a + b)|
- The lines are perpendicular when a + b = 0 (coefficient of x2 + coefficient of y2 = 0).
- The lines are coincident (parallel) when h2 = ab.
The full second-degree equation represents a pair of straight lines only if its determinant condition holds: abc + 2fgh − af2 − bg2 − ch2 = 0. For NDA, the two boxed results above (a + b = 0 and h2 = ab) are the ones asked most often.
Worked Example - Line Through Two Points
Find the equation of the line passing through A(2, 3) and B(5, 9), and then find its perpendicular distance from the origin.
So the line is 2x − y − 1 = 0 and its distance from the origin is 1 ÷ √5 (about 0.447 units). Notice how three formulas chained together — slope, point-slope, then perpendicular distance — solved the whole problem cleanly.
Common Mistakes to Avoid
This chapter is a scoring one, but careless slips quietly cost marks. Guard against these traps.
- Writing the slope as (x2 − x1) ÷ (y2 − y1) — it is (y2 − y1) ÷ (x2 − x1), rise over run.
- Forgetting that a vertical line has an undefined slope, not a slope of zero.
- Using the condition for parallel (m1 = m2) when the lines are perpendicular — perpendicular needs m1m2 = −1.
- Dropping the modulus in the distance formula and getting a negative distance.
- Reading a and b in intercept form as coefficients instead of actual intercepts.
For a pair of lines through the origin, students mix up the sum and product of slopes. Remember: m1 + m2 = −2h ÷ b (note the minus and the 2), while m1m2 = a ÷ b.
Previous-Year Style Practice
Let us apply the pair-of-lines results to a question in the exact style the NDA loves.
Q. The equation 2x2 + 7xy + 3y2 = 0 represents a pair of straight lines through the origin. What is the angle between them, and are they perpendicular?
Answer: Here a = 2, 2h = 7 so h = 7÷2, and b = 3. The lines are perpendicular only if a + b = 0; but a + b = 2 + 3 = 5 ≠ 0, so they are not perpendicular. The angle is tan θ = |2√(h2 − ab) ÷ (a + b)| = |2√(49÷4 − 6) ÷ 5| = |2√(25÷4) ÷ 5| = |2 × (5÷2) ÷ 5| = |5 ÷ 5| = 1. So tan θ = 1, giving θ = 45°.
Notice the workflow: read off a, h, b; test the perpendicular condition a + b = 0 first; then plug into the angle formula. With practice this whole question takes under a minute.
Quick Revision
- A line is a first-degree equation: ax + by + c = 0; its slope is m = −a ÷ b.
- Slope = (y2 − y1) ÷ (x2 − x1) = tan θ.
- Parallel: m1 = m2; Perpendicular: m1m2 = −1.
- Distance of (x1, y1) from ax + by + c = 0 = |ax1 + by1 + c| ÷ √(a2 + b2).
- Angle: tan θ = |(m1 − m2) ÷ (1 + m1m2)|.
- Pair of lines ax2 + 2hxy + by2 = 0: perpendicular if a + b = 0, coincident if h2 = ab.
- Three collinear points give triangle area zero.
Drill these seven formulas the night before your exam and the straight-line questions become almost automatic marks.
Frequently asked questions
How many questions on straight lines come in the NDA Maths exam?
Coordinate geometry, with straight lines at its core, typically contributes 3 to 5 questions in each NDA Maths paper. They are short and formula-based, covering slope, line equations, distance of a point, angle between lines and the pair of straight lines, making this a high-scoring chapter.
What is the slope of a line given as ax + by + c = 0?
The slope is m = −a ÷ b, that is, minus the coefficient of x divided by the coefficient of y. You can read it straight off the equation without rearranging into y = mx + c form, which saves time in the exam.
How do I check whether two lines are perpendicular?
Two lines are perpendicular when the product of their slopes equals −1, that is m1 × m2 = −1. For a pair of lines given by ax² + 2hxy + by² = 0, they are perpendicular when a + b = 0, meaning the coefficients of x² and y² add to zero.
What is the formula for the perpendicular distance of a point from a line?
For a point (x1, y1) and a line ax + by + c = 0, the perpendicular distance is d = |ax1 + by1 + c| divided by the square root of (a² + b²). The modulus keeps the distance positive, so always include it.
When does a second-degree equation represent a pair of straight lines?
The homogeneous equation ax² + 2hxy + by² = 0 always gives two straight lines through the origin. The full equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of lines only when abc + 2fgh − af² − bg² − ch² = 0.
How do I find the angle between two straight lines?
If the slopes are m1 and m2, then tan θ = |(m1 − m2) ÷ (1 + m1m2)|. The lines are parallel when m1 = m2 (angle 0°) and perpendicular when 1 + m1m2 = 0 (angle 90°).
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