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Straight Lines and Cartesian System hero

Straight Lines and Cartesian System

~14 min read

In 30 seconds
  • What: Straight Lines and Cartesian System covers distance, slope, various line equations, angle between lines, perpendicular distance, and locus problems in a 2D coordinate plane.
  • Why it matters: This is one of the highest-frequency topics in NDA Mathematics — questions appear every year from 2010 to 2025, often as 6–10 questions per paper.
  • Key fact: Mastering the perpendicular-distance formula and conditions for parallel/perpendicular lines alone can secure 4–6 marks per paper.

The Cartesian coordinate system gives you a way to pin any point in a plane using two numbers. Once you understand that, straight lines become easy to handle — you describe them with equations, measure distances between them, find where they cross, and work out the angles they make. NDA papers have tested every variant of this topic since at least 2010, making it one of the safest areas to invest preparation time in.

What This Topic Covers

This topic sits at the intersection of algebra and geometry. The Cartesian plane — two perpendicular number lines meeting at the origin — is your workspace. Every concept listed below appears in NDA PYQs, some more than once per paper.

Subtopics tested in NDA

  • Distance between two points and section formula (internal and external division)
  • Collinearity of three points and area of a triangle with given vertices
  • Slope of a line; conditions for parallel and perpendicular lines
  • Various forms of line equations: slope-intercept, point-slope, two-point, intercept, normal
  • Distance from a point to a line; distance between parallel lines
  • Angle between two lines
  • Locus problems — equidistant points, perpendicular bisectors
  • Centroid, circumcentre, incentre, orthocentre of a triangle
  • Concurrent lines and family of lines through intersection points

Questions on this topic also connect to Circles and Conic Sections and Three-Dimensional Geometry, so a strong base here pays off across multiple chapters.

Exam Pattern & Weightage

The table below summarises question counts drawn directly from the PYQ file. Years shown are those explicitly cited in the source questions.

Year Paper No. Key Sub-areas
2010 I & II 5 Intersection of lines, slope, area of triangle, perpendicular lines
2011 I & II 7 Locus, foot of perpendicular, parallel lines, collinearity
2012 I & II 8 Perpendicular distance, intercepts, angle of inclination, parallel lines
2013 I & II 9 Passing through intersection, concurrent lines, centroid, perimeter
2014 I & II 5 Bisected segment, equal intercepts, collinear condition
2015 I & II 9 Normal form, circumcentre, concurrent lines, perpendicular distance
2016 I & II 10 Area of triangle/parallelogram, midpoint, intersection, diagonal of square
2017 I & II 7 Parallel lines, centroid, collinear points, intercept form, distance
2018 I & II 9 Distance between parallel lines, angle between lines, intercept divided
2019 I & II 7 Perpendicular from origin, locus, equidistant point, concurrent
2020 I 6 Diagonal of square, circumcentre, collinear condition, area of square
2021 I & II 8 Fixed point on family, image in line mirror, diagonal of parallelogram
2022 I & II 6 Equal intercepts, parallel conditions, locus of midpoint
2023 I & II 5 Normal form sum of intercepts, section ratio, parallel lines condition
2024 I & II 7 Centroid, isosceles triangle, perpendicular to line, rhombus/rectangle
2025 I 5 Perpendicular condition, equilateral triangle locus, angular bisectors
NDA Alert

In 2016, this single topic contributed 10 questions across the two papers — more than any other coordinate geometry subtopic. Never skip this chapter in your revision.

Core Concepts

Distance and Section Formulas

The distance between two points is the most fundamental building block. The section formula tells you the coordinates of a point dividing a segment in a given ratio.

Distance Formula Distance between \((x_1, y_1)\) and \((x_2, y_2)\): $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Internal Division (ratio m:n) $$P = \left( \frac{mx_2 + nx_1}{m+n},\ \frac{my_2 + ny_1}{m+n} \right)$$
External Division (ratio m:n) $$P = \left( \frac{mx_2 - nx_1}{m-n},\ \frac{my_2 - ny_1}{m-n} \right)$$
Midpoint $$M = \left( \frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2} \right)$$
Common Trap

The external-division formula uses minus signs in both numerator and denominator. If the question says "P divides AB externally in ratio \(3:2\)," do not reuse the internal formula — sign errors here cost the whole mark. Also, for external division the point P lies outside the segment AB.

Slope and Conditions for Parallel / Perpendicular Lines

The slope \(m\) of a line measures its steepness. It links directly to whether two lines are parallel or perpendicular — questions that appear almost every year.

Slope of a Line $$m = \frac{y_2 - y_1}{x_2 - x_1} = \tan \theta$$ where \(\theta\) is the angle the line makes with the positive x-axis.
Slope from General Form \(ax + by + c = 0\) $$m = -\frac{a}{b} = -\frac{\text{coefficient of } x}{\text{coefficient of } y}$$
Common Trap

For a vertical line (parallel to the y-axis), the slope is undefined — not "infinity." The denominator \(x_2 - x_1\) is zero, so \(\tan 90^\circ\) does not exist. NDA answer keys explicitly mark "undefined" or "not defined"; avoid choosing "∞" if it appears as an option.

Parallel Lines $$m_1 = m_2$$ (slopes are equal). For \(ax + by + c = 0\) and \(ax + by + d = 0\): coefficients of \(x\) and \(y\) in the same ratio.
Perpendicular Lines $$m_1 \cdot m_2 = -1$$ For \(ax + by + c = 0\), any perpendicular line has the form \(bx - ay + k = 0\) (swap coefficients, flip the middle sign).
Common Trap

The perpendicular condition is the product of slopes \(= -1\), not the sum. Students often write \(m_1 + m_2 = 0\) by reflex — that is the condition for slopes of equal magnitude opposite sign, not perpendicularity. Also, the product rule fails when one line is vertical (slope undefined); in that case the perpendicular is simply horizontal (\(m = 0\)).

The 2011 PYQ asked: "Two straight lines \(x - 3y - 2 = 0\) and \(2x - 6y - 6 = 0\) — do they intersect?" Comparing ratios of coefficients: \(\tfrac{1}{2} = \tfrac{-3}{-6} = \tfrac{1}{2}\), but \(\tfrac{-2}{-6} \ne \tfrac{1}{2}\), so the lines are parallel and never intersect. This ratio-check is a 30-second technique for parallel-line questions.

Forms of the Equation of a Straight Line

Slope-Intercept Form $$y = mx + c$$ (\(m\) = slope, \(c\) = y-intercept)
Point-Slope Form $$y - y_1 = m(x - x_1)$$
Two-Point Form $$\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$$
Intercept Form $$\frac{x}{a} + \frac{y}{b} = 1$$ (\(a\) = x-intercept, \(b\) = y-intercept). Reciprocals of intercepts \(m\) and \(n\) give: \(mx + ny = 1\).
Normal (Perpendicular) Form $$x \cos \alpha + y \sin \alpha = p$$ where \(p\) is the perpendicular distance from the origin and \(\alpha\) is the angle the normal makes with the positive x-axis.
General Form $$ax + by + c = 0$$ represents a line when at least one of \(a, b\) is non-zero (NDA 2019-II).
NDA Alert

A 2012 PYQ asked: "The equation of a straight line which makes an angle \(45^\circ\) with the x-axis with y-intercept \(101\) units." Answer: \(y = x + 101\), i.e., \(x - y + 101 = 0\). The slope is \(\tan 45^\circ = 1\). Always link angle of inclination to \(\tan \theta\) immediately.

Angle Between Two Lines

Angle Between Two Lines $$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$ The modulus guarantees the acute angle; the obtuse angle is \(180^\circ - \theta\).

A 2013 PYQ asked for the angle between \(x + y = 1\) and \(x - y = 1\). Slopes are \(-1\) and \(1\). \(\tan \theta = \left|\tfrac{-1-1}{1+(-1)(1)}\right| = \left|\tfrac{-2}{0}\right|\) → undefined, so \(\theta = 90^\circ\). The lines are perpendicular.

Perpendicular Distance

Distance from Point \((x_1, y_1)\) to Line \(ax + by + c = 0\) $$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$
Distance Between Parallel Lines \(ax + by + c_1 = 0\) and \(ax + by + c_2 = 0\) $$d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$$ Warning: the \(x\)- and \(y\)-coefficients of both lines must match exactly before subtracting the constants. Re-scale first if they don't.

A 2012 PYQ: "Distance between \(3x + 4y = 9\) and \(-6x - 8y = 18\)." Rewrite second line: \(3x + 4y + 9 = 0\). Distance \(= |9 - (-9)| / \sqrt{9+16} = 18/5\) — but the answer was \(0\) because rewriting: \(6x + 8y = 18\) simplifies to \(3x + 4y = 9\), which is the same line. Another 2018 PYQ: distance between \(3x + 4y = 9\) and \(6x + 8y = 15\). Rewrite second: \(3x + 4y = 7.5\). Distance \(= |9 - 7.5|/5 = 1.5/5 = 3/10\).

Perpendicular from Origin to \(\tfrac{x}{a} + \tfrac{y}{b} = 1\) $$\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$$

Area of Triangle and Collinearity

Area of Triangle with Vertices \((x_1,y_1)\), \((x_2,y_2)\), \((x_3,y_3)\) $$\text{Area} = \tfrac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$
Collinearity Condition Three points are collinear when the above area \(= 0\).

A 2010 PYQ: area of triangle with vertices \((-3,0)\), \((3,0)\), \((0,k)\) \(= 9\) sq units. Using the formula: \(\tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 6 \times |k| = 9\), so \(|k| = 3\).

Special Points of a Triangle

Centroid $$G = \left( \frac{x_1 + x_2 + x_3}{3},\ \frac{y_1 + y_2 + y_3}{3} \right)$$

Centroid questions appear heavily — 2013, 2017, 2019, 2024 all have at least one centroid PYQ. Given three vertices or midpoints, you will need to reconstruct vertices and then find the centroid.

Locus Problems

A locus is the path traced by a point satisfying a given condition. NDA tests two recurring locus types:

  • Equidistant from two fixed points: The locus is the perpendicular bisector of the segment joining those two points.
  • Equidistant from the coordinate axes: The locus is \(y = \pm x\), i.e., the equations \(x - y = 0\) or \(x + y = 0\) (2011 PYQ).
Locus Equidistant from \((a,0)\) and \((0,b)\) Set distance from \((x,y)\) to \((a,0)\) equal to distance from \((x,y)\) to \((0,b)\); simplify to get the equation of the perpendicular bisector.

Family of Lines Through Intersection

Family of Lines Through Intersection of \(L_1 = 0\) and \(L_2 = 0\) $$L_1 + \lambda L_2 = 0$$ for varying values of \(\lambda\). Use the extra condition (parallel, passing through a point, etc.) to find \(\lambda\).

This technique appears in 2013, 2017, 2018, and 2019 PYQs. You are given two intersecting lines and asked for the equation of a third line through their intersection satisfying another condition.

Worked Examples

Example 1 — Parallel Lines Check (2011)

Q: Two straight lines \(x - 3y - 2 = 0\) and \(2x - 6y - 6 = 0\) — what is the relationship?

  • Compare ratios of coefficients: \(\tfrac{a_1}{a_2} = \tfrac{1}{2}\), \(\tfrac{b_1}{b_2} = \tfrac{-3}{-6} = \tfrac{1}{2}\). Since \(\tfrac{a_1}{a_2} = \tfrac{b_1}{b_2}\), the lines are parallel or coincident.
  • Check \(c\) ratio: \(\tfrac{c_1}{c_2} = \tfrac{-2}{-6} = \tfrac{1}{3} \ne \tfrac{1}{2}\). So the lines are parallel but not coincident.
  • Answer: the lines never intersect.

Example 2 — Distance Between Parallel Lines (2015)

Q: Perpendicular distance between \(6x + 8y + 15 = 0\) and \(3x + 4y + 9 = 0\).

  • Make coefficients of \(x\) and \(y\) equal: multiply the second line by \(2\) to get \(6x + 8y + 18 = 0\).
  • Now both lines have the form \(6x + 8y + c = 0\), with \(c_1 = 15\) and \(c_2 = 18\).
  • Distance: $$d = \frac{|15 - 18|}{\sqrt{36 + 64}} = \frac{3}{\sqrt{100}} = \frac{3}{10}\ \text{units}.$$

Example 3 — Line Through Intersection, Parallel Condition (2018)

Q: Equation of line through intersection of \(\tfrac{x}{2} + \tfrac{y}{3} = 1\) and \(\tfrac{x}{3} + \tfrac{y}{2} = 1\), parallel to \(4x + 5y - 6 = 0\).

  • Find intersection: rewrite as \(3x + 2y = 6\) and \(2x + 3y = 6\). Subtracting: \(x - y = 0\), so \(x = y\). Substituting: \(5x = 6\), \(x = \tfrac{6}{5}\). Point \(= \left(\tfrac{6}{5}, \tfrac{6}{5}\right)\).
  • Required line is parallel to \(4x + 5y - 6 = 0\), so its form is \(4x + 5y + k = 0\).
  • Substitute \(\left(\tfrac{6}{5}, \tfrac{6}{5}\right)\): \(4 \cdot \tfrac{6}{5} + 5 \cdot \tfrac{6}{5} + k = 0\) \(\implies \tfrac{24}{5} + \tfrac{30}{5} + k = 0\) \(\implies k = -\tfrac{54}{5}\).
  • Multiply through by \(5\): \(20x + 25y - 54 = 0\). Answer: \(20x + 25y - 54 = 0\).

Example 4 — Locus Equidistant from Two Points (2013)

Q: A point \(P\) moves equidistant from \((1, 2)\) and \((-2, 3)\). Find the locus.

  • Set \(PA = PB\): $$\sqrt{(x-1)^2 + (y-2)^2} = \sqrt{(x+2)^2 + (y-3)^2}.$$
  • Square both sides: \((x-1)^2 + (y-2)^2 = (x+2)^2 + (y-3)^2\).
  • Expand: \(x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 + 4x + 4 + y^2 - 6y + 9\).
  • Simplify: \(-2x + 1 - 4y + 4 = 4x + 4 - 6y + 9\) \(\implies -6x + 2y - 8 = 0\) \(\implies 3x - y + 4 = 0\). This is a straight line (perpendicular bisector).

Example 5 — Perpendicular Line via Coefficient Swap

Q: Find the equation of the line passing through \((2, 3)\) and perpendicular to \(4x + 3y - 10 = 0\).

  • Use the shortcut: any line perpendicular to \(ax + by + c = 0\) has the form \(bx - ay + k = 0\). Swap coefficients, flip the middle sign.
  • Required line: \(3x - 4y + k = 0\).
  • Substitute \((2, 3)\): \(3(2) - 4(3) + k = 0\) \(\implies 6 - 12 + k = 0\) \(\implies k = 6\).
  • Answer: \(3x - 4y + 6 = 0\). (No slope calculation needed.)

Example 6 — Perpendicular Distance from a Point

Q: Find the perpendicular distance from the point \((3, -4)\) to the line \(3x - 4y + 7 = 0\).

  • Apply $$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$ with \((x_1, y_1) = (3, -4)\), \(a = 3\), \(b = -4\), \(c = 7\).
  • Numerator: \(|3(3) + (-4)(-4) + 7| = |9 + 16 + 7| = 32\).
  • Denominator: \(\sqrt{9 + 16} = \sqrt{25} = 5\).
  • Answer: \(d = \tfrac{32}{5} = 6.4\) units.

Example 7 — Area of Triangle with Given Vertices (2013)

Q: Area of the triangle with vertices \((3, 0)\), \((0, 4)\) and \((3, 4)\).

  • Apply area formula: \(\tfrac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|\).
  • \(= \tfrac{1}{2} |3(4-4) + 0(4-0) + 3(0-4)|\).
  • \(= \tfrac{1}{2} |3 \times 0 + 0 + 3 \times (-4)| = \tfrac{1}{2} \times 12 = 6\) sq. units.

How NDA Tests This Topic

NDA rarely asks straightforward formula-plug questions. Instead, the paper combines two ideas in one question — for example, finding the line through the intersection of two given lines (family of lines) that is also parallel to a third line. Practising each such combination separately is the most efficient preparation approach.

Test Your Straight Lines Skills

Attempt NDA-pattern mock questions on this topic — perpendicular distance, locus, and family of lines — timed to build exam-hall speed.

Start Free Mock Test

Exam Shortcuts (Pro-Tips)

Straight Lines questions reward speed. The shortcuts below collapse 1–2 minutes of algebra into 10–15 seconds each. Every one of them has shown up in an NDA paper — memorise them cold.

Shortcut 1 — Perpendicular Slope Product = −1

For any two non-vertical lines with slopes \(m_1\) and \(m_2\), perpendicularity means \(m_1 \cdot m_2 = -1\). The direct corollary for general-form lines \(a_1 x + b_1 y + c_1 = 0\) and \(a_2 x + b_2 y + c_2 = 0\) is:

Perpendicularity (General Form) $$a_1 a_2 + b_1 b_2 = 0$$

No need to compute slopes — multiply the x-coefficients, multiply the y-coefficients, add. Sum = 0 means perpendicular. This shortcut alone solves the 2025-I perpendicular-condition question in under 10 seconds.

Shortcut 2 — Parallel Lines Share the Coefficient Block

Any line parallel to \(ax + by + c = 0\) has the form \(ax + by + k = 0\) — only the constant changes. So if a question asks "line through \((x_0, y_0)\) parallel to \(3x - 4y + 7 = 0\)," write \(3x - 4y + k = 0\) immediately, substitute the point, solve for \(k\). Skip slope formulas entirely.

Shortcut 3 — Distance from Origin

For the line \(ax + by + c = 0\), set \((x_1, y_1) = (0, 0)\) in the perpendicular distance formula:

Distance from Origin to \(ax + by + c = 0\) $$d = \frac{|c|}{\sqrt{a^2 + b^2}}$$

NDA 2019-I asked "perpendicular distance from origin to \(3x + 4y - 25 = 0\)" — answer \(= |-25|/\sqrt{9+16} = 25/5 = 5\) in one step.

Shortcut 4 — Midpoint = Section Formula at 1:1

The midpoint \(M\) of \((x_1, y_1)\) and \((x_2, y_2)\) is just the section formula with ratio \(1:1\), giving $$M = \left( \frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2} \right).$$ Use this everywhere — diagonals of a parallelogram bisect each other (so the midpoints of both diagonals coincide), and the circumcentre of a right triangle is the midpoint of the hypotenuse.

Shortcut 5 — Section Formula: Sign Flip for External Division

Internal division (point \(P\) lies between \(A\) and \(B\)): plus signs. External division (\(P\) lies outside segment \(AB\)): minus signs in numerator and denominator.

External Division (ratio \(m:n\)) $$P = \left( \frac{mx_2 - nx_1}{m-n},\ \frac{my_2 - ny_1}{m-n} \right)$$

Quick check: external division in ratio \(m:n\) is the same as internal division in ratio \(m:(-n)\). If you ever forget the formula, just plug a negative ratio into the internal formula.

Shortcut 6 — Area of Triangle via Coordinates (Determinant Form)

Area of Triangle with Vertices \((x_1,y_1)\), \((x_2,y_2)\), \((x_3,y_3)\) $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$

Setting Area \(= 0\) instantly gives the collinearity test. For a triangle with one side on the x-axis or y-axis, the formula collapses to \(\tfrac{1}{2} \times \text{base} \times \text{height}\) — even faster.

Shortcut 7 — Foot of Perpendicular and Image Direct Formulas

For line \(ax + by + c = 0\) and point \((x_1, y_1)\), let \((h, k)\) be the foot of the perpendicular and \((x', y')\) be the image (reflection):

Foot of Perpendicular $$\frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$$
Image of Point (Reflection) $$\frac{x' - x_1}{a} = \frac{y' - y_1}{b} = -2 \cdot \frac{ax_1 + by_1 + c}{a^2 + b^2}$$

The image formula is literally twice the foot-of-perpendicular value — same denominator, multiply the right side by \(2\). The 2021 PYQ on "image of a point in a line mirror" is a 20-second problem with this.

Shortcut 8 — Concurrency of Three Lines (Determinant Test)

Three lines \(a_1 x + b_1 y + c_1 = 0\), \(a_2 x + b_2 y + c_2 = 0\), \(a_3 x + b_3 y + c_3 = 0\) are concurrent (meet at one point) iff:

Concurrency Condition $$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$$

Used in 2013-I and 2015-II concurrency questions. Beats solving two equations and substituting into the third.

Shortcut 9 — Right-Triangle Orthocentre and Circumcentre

  • In a right-angled triangle, the orthocentre sits exactly at the vertex containing the \(90^\circ\) angle.
  • The circumcentre sits exactly at the midpoint of the hypotenuse.
  • For the triangle formed by the x-axis, y-axis, and the line \(\tfrac{x}{a} + \tfrac{y}{b} = 1\), the right angle is at the origin — so orthocentre \(= (0, 0)\) instantly, and circumcentre \(= \left(\tfrac{a}{2}, \tfrac{b}{2}\right)\).

Common Question Patterns

After reviewing PYQs from 2010 to 2025, these are the question types that recur most frequently:

Pattern 1 — Parallel/Perpendicular Condition

  • Given two lines \(ax + by + c = 0\) and \(kx + dy + e = 0\), find \(k\) for parallel or perpendicular.
  • Key rule: parallel \(\Leftrightarrow \tfrac{a}{k} = \tfrac{b}{d} \ne \tfrac{c}{e}\); perpendicular \(\Leftrightarrow\) product of slopes \(= -1\), i.e., \(a \cdot k + b \cdot d = 0\).
  • Appeared: 2011-II, 2017-I, 2021-I, 2022-I, 2025-I.

Pattern 2 — Distance from Point to Line or Between Parallel Lines

  • Asked directly or embedded in finding the area of a triangle with one vertex and one side given.
  • Appeared: 2012-I, 2015-I, 2018-I, 2019-I, 2023-I.

Pattern 3 — Centroid / Circumcentre of a Triangle

  • Vertices given directly, or one vertex and two midpoints given — reconstruct missing vertex, then find centroid.
  • Centroid: average of coordinates. Circumcentre: equidistant from all three vertices.
  • Appeared: 2013-II, 2015-I, 2017-I, 2019-I, 2021-II, 2024-I.

Pattern 4 — Line Through Intersection of Two Lines

  • Use family of lines: \(L_1 + \lambda L_2 = 0\), then apply the extra condition (parallel, perpendicular, passes through a point).
  • Appeared: 2013-I, 2017-II, 2018-II, 2019-I.

Pattern 5 — Locus Problems

  • Set up the condition algebraically, square both sides if needed, and simplify.
  • Common results: perpendicular bisector (straight line), or \(y = \pm x\) for equidistant from axes.
  • Appeared: 2011-I, 2012-II, 2013-II, 2022-I, 2025-I.

Pattern 6 — Intercept Form and Normal Form

  • Given intercepts sum, product, or ratio — or given perpendicular distance and angle of normal — write the line equation.
  • Appeared: 2013-II, 2015-I, 2017-II, 2023-I.
NDA Alert

When the question gives "the line \(ax + by + c = 0\)" and asks what value of \(k\) makes \(kx + by + c = 0\) parallel, always check whether \(a = k\) is the only solution or whether \(a = -k\) (for perpendicular) is being asked instead. Misreading the condition is the most common slip on this question type.

Preparation Strategy

This topic has enough sub-areas that unfocused revision loses you time. Here is a structured approach:

Week 1 — Formulas and Basic Application

  • Write every formula in this page on one A4 sheet — distance, section, slope, all line forms, perpendicular distance, area.
  • Solve 20 straight computation questions (no tricks, just formula application). PYQs from 2010–2012 are ideal for this.
  • Target: zero formula errors before moving on.

Week 2 — Combination and Multi-Step Questions

  • Practise family-of-lines questions. Given two intersecting lines, find the third line through their intersection satisfying a condition.
  • Practise centroid, circumcentre, and orthocentre identification — they frequently appear together in a single 3-question set (as in 2015 PYQ).
  • Work through all locus problems from 2011–2023 without looking at options first.

Week 3 — Timed Mock Practice

  • Take 10-question timed sets from this chapter alone. Each question must be solved within 90 seconds.
  • After each set, identify which sub-area caused errors and revise only that concept.
  • Attempt full NDA mock tests and track this chapter's score separately.

Connected Topics to Cover Together

High-Value Shortcuts to Memorise

  • For lines \(px + qy + r = 0\) and \(qx + py + r = 0\): they are parallel when \(p = q\); perpendicular when \(p^2 = q^2\) i.e. \(p = \pm q\), with the \(+\) case giving coincident lines (check 2022 PYQ).
  • When \(A, B, C\) are in AP, the line \(Ax + 2By + C = 0\) always passes through \((1, -1)\) (NDA 2021-I).
  • Line \(y = 0\) divides segment joining \((x_1, y_1)\) and \((x_2, y_2)\) in ratio \(|y_1| : |y_2|\) (external if signs are same).
  • The locus of a point equidistant from three collinear points is the null set — there is no such point (2012 PYQ).

Track your accuracy chapter-by-chapter using the full NDA Maths subject index to see which areas still need work. Pair this with Matrices and Determinants revision, as the determinant formula for area of a triangle is the same computation in matrix form.

Frequently Asked Questions

How many questions come from Straight Lines in each NDA paper?

Based on PYQs from 2010 to 2025, the count ranges from 5 to 10 questions per exam (both papers combined). The topic is consistently one of the top contributors in the coordinate geometry section of the NDA Mathematics paper.

What is the most important formula for this topic?

The perpendicular distance formula — $$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$ — and its partner for distance between parallel lines are the most frequently tested. Master these first, then the parallel/perpendicular slope conditions.

What does the equation ax + by + c = 0 represent?

It represents a straight line whenever at least one of \(a\) or \(b\) is non-zero. If both \(a\) and \(b\) are zero, it is not a line equation. This exact point was tested in NDA 2019-II.

How do I find the line through the intersection of two lines satisfying an extra condition?

Use the family of lines formula: \(L_1 + \lambda L_2 = 0\). Substitute the extra condition (e.g., parallel slope, a specific point) to find \(\lambda\), then write the final equation. This avoids solving for the intersection point first.

What is the locus of a point equidistant from both coordinate axes?

The locus is \(y = x\) or \(y = -x\), i.e., the two angle bisectors of the axes. In equation form: \(x - y = 0\) and \(x + y = 0\). This was asked in NDA 2011-I and again in 2013-II.

How is the centroid different from the circumcentre?

The centroid is the average of the three vertex coordinates and always lies inside the triangle. The circumcentre is equidistant from all three vertices and can lie outside the triangle for obtuse triangles. NDA has asked for both in separate questions within the same paper (see the 2015-I set on triangle \(ABC\)).

When A, B, C are in AP, what fixed point does Ax + 2By + C = 0 pass through?

When \(A, B, C\) are in arithmetic progression, the line \(Ax + 2By + C = 0\) always passes through the fixed point \((1, -1)\). This was directly asked in NDA 2021-I. Verify by substituting: \(A(1) + 2B(-1) + C = A - 2B + C = 0\) when \(2B = A + C\), which is the AP condition.