Geometry and coordinate basics

In short

Geometry in aptitude tests is the study of angles, triangles, quadrilaterals, circles and polygons through a small set of standard theorems, while coordinate basics place the same figures on the x-y plane so that distances, slopes and areas can be computed by formula instead of construction. The questions are short and rule-driven: identify the configuration, apply the right theorem, finish in 60-90 seconds. It is examined in CAT, XAT, CMAT, SSC CGL/CHSL, RRB, IBPS and SBI PO, the GATE General Aptitude section, and in company aptitude rounds (TCS NQT, Infosys, Wipro, Capgemini, Accenture, Cognizant, L&T, BEL, ISRO/DRDO written tests). The same formulas are the working arithmetic of civil surveying, CAD/CAM, GIS, robotics and computer graphics, so interviewers in those fields treat them as assumed knowledge.

Explained

Lines, angles and polygons

• Angle types: acute (< 90°), right (90°), obtuse (90°-180°), straight (180°), reflex (180°-360°). Complementary angles sum to 90°, supplementary to 180°.
• At a point on a line, angles on one side sum to 180° (linear pair); around a point they sum to 360°; vertically opposite angles are equal.
• Two parallel lines cut by a transversal: corresponding angles equal, alternate interior angles equal, co-interior (allied) angles supplementary. Converses are used to *prove* lines parallel.
• Polygon (n sides): sum of interior angles = (n − 2) × 180°; sum of exterior angles = 360° always; number of diagonals = n(n − 3)/2.
• Regular polygon: each interior angle = 180° − 360°/n; each exterior angle = 360°/n. A regular polygon tiles the plane only if the interior angle divides 360° (triangle, square, hexagon).
• Convex vs concave: every interior angle < 180° in a convex polygon; one reflex angle makes it concave.

Triangles - the core of the topic

• Angle sum = 180°. Exterior angle = sum of the two remote interior angles.
• Triangle inequality: any side < sum of the other two and > their difference. So sides 3, 4, 8 cannot form a triangle.
• Classification: by sides - scalene, isosceles (base angles equal), equilateral (all 60°); by angles - acute, right, obtuse.
• Pythagoras theorem: in a right triangle, hypotenuse² = sum of squares of the legs. Converse identifies right triangles. If c² > a² + b² the triangle is obtuse, if less, acute.
• Pythagorean triplets to memorise: 3-4-5, 5-12-13, 8-15-17, 7-24-25, 9-40-41, 20-21-29, and all their multiples (6-8-10, 9-12-15, 10-24-26).
• Congruence (same shape and size): SSS, SAS, ASA, AAS, and RHS for right triangles. AAA is not congruence - it is similarity.
• Similarity (same shape): AA (or AAA), SSS (sides in proportion), SAS (two sides proportional with the included angle equal). In similar triangles corresponding sides and perimeters are in ratio k, areas in ratio k².
• Basic Proportionality Theorem (Thales/BPT): a line parallel to one side divides the other two proportionally - DE ∥ BC ⇒ AD/DB = AE/EC. Converse holds.
• Mid-point theorem: the segment joining mid-points of two sides is parallel to the third side and half its length.
• Angle bisector theorem: the internal bisector from A meets BC at D with BD/DC = AB/AC.
• Right-triangle altitude relations: the altitude to the hypotenuse creates two triangles similar to the original; h = ab/c, and each leg² = hypotenuse × its adjacent segment.

Triangle centres and standard lengths

• Centroid (G): intersection of the three medians; divides each median 2:1 from the vertex; it is the centre of mass of a uniform triangular plate.
• Incentre (I): intersection of internal angle bisectors; centre of the inscribed circle, r = Area/s where s is the semi-perimeter.
• Circumcentre (O): intersection of perpendicular bisectors of the sides; R = abc/(4 × Area). In a right triangle it is the mid-point of the hypotenuse, so R = hypotenuse/2; it lies outside an obtuse triangle.
• Orthocentre (H): intersection of the altitudes; it is the right-angle vertex in a right triangle.
• O, G and H are collinear (Euler line) with OG : GH = 1 : 2. In an equilateral triangle all four centres coincide.
• Median length (Apollonius): 4m_a² = 2b² + 2c² − a².
• Equilateral triangle of side a: height = (√3/2)a, R = a/√3, r = a/(2√3), so R = 2r.

Quadrilaterals

• Interior angles of any quadrilateral sum to 360°.
• Parallelogram law: sum of squares of the diagonals = 2(a² + b²).
• The mid-points of the sides of *any* quadrilateral form a parallelogram (Varignon).
• A quadrilateral is cyclic (vertices on one circle) if and only if opposite angles are supplementary; an exterior angle then equals the opposite interior angle.

Circles

• Terms: centre, radius, chord, diameter (longest chord), arc, sector, segment, secant (cuts at two points), tangent (touches at one point).
• Chords: a perpendicular from the centre bisects the chord, and conversely; equal chords are equidistant from the centre; the chord at distance d from the centre has length 2√(r² − d²).
• Angles: the angle subtended at the centre = twice the angle at any point on the remaining circumference; angles in the same segment are equal; the angle in a semicircle is 90°.
• Tangents: a tangent is perpendicular to the radius at the point of contact; the two tangents from an external point are equal in length and subtend equal angles at the centre.
• Alternate segment theorem: the angle between a tangent and a chord equals the inscribed angle in the alternate segment.
• Power of a point: two chords meeting inside at P give PA·PB = PC·PD; two secants from an outside point give PA·PB = PC·PD; and tangent² PT² = PA·PB.
• Two circles: distance between centres D with radii r₁, r₂ - D > r₁ + r₂ (separate, 4 common tangents), D = r₁ + r₂ (externally tangent, 3), |r₁ − r₂| < D < r₁ + r₂ (intersecting, 2), D = |r₁ − r₂| (internally tangent, 1), D < |r₁ − r₂| (one inside the other, 0).
• Direct common tangent length = √(D² − (r₁ − r₂)²); transverse common tangent length = √(D² − (r₁ + r₂)²).

Coordinate geometry - points and lines

• Axes and quadrants: signs (+,+), (−,+), (−,−), (+,−) in quadrants I to IV. Points on the x-axis have y = 0; on the y-axis x = 0.
• Distance, mid-point, section formula are the three workhorses (boxed below). The external section formula uses m − n in the denominator.
• Centroid of a triangle = average of the three vertices. The in-centre is weighted by side lengths: (ax₁ + bx₂ + cx₃)/(a+b+c), similarly for y.
• Slope m = tan θ = (y₂ − y₁)/(x₂ − x₁). A horizontal line has m = 0; a vertical line has undefined slope. Parallel ⇒ m₁ = m₂; perpendicular ⇒ m₁m₂ = −1.
• Forms of a line: slope-intercept y = mx + c; point-slope y − y₁ = m(x − x₁); two-point; intercept x/a + y/b = 1; general ax + by + c = 0 with slope −a/b, x-intercept −c/a, y-intercept −c/b.
• Collinearity: three points are collinear if the area of the triangle they form is zero, or if the slopes of two pairs are equal.
• Angle between lines, distance from a point to a line, distance between parallel lines - all boxed below.
• Reflections: over the x-axis (x, −y); y-axis (−x, y); origin (−x, −y); y = x gives (y, x); y = −x gives (−y, −x). Reflection is the standard trick for "minimise PA + PB with P on a line" problems.
• Translation shifts by (h, k); rotation of 90° anticlockwise about the origin sends (x, y) to (−y, x).
• Circle: (x − h)² + (y − k)² = r², or x² + y² + 2gx + 2fy + c = 0 with centre (−g, −f) and radius √(g² + f² − c). The equation represents a real circle only if g² + f² − c > 0; a line is a tangent when the perpendicular distance from the centre equals r.
• Three dimensions (GATE/engineering context): distance = √(Δx² + Δy² + Δz²); the same section and mid-point formulas apply with a third coordinate.

Key points

• Angle sum - 180° in a triangle, 360° in a quadrilateral, (n − 2)180° in an n-gon.
• Exterior angles - of any convex polygon always total 360°, whatever n.
• Triangle inequality - each side lies strictly between the sum and the difference of the other two.
• AAA - proves similarity, never congruence; RHS applies only to right triangles.
• Area ratio - in similar figures it is the square of the linear ratio; volumes scale as the cube.
• Centroid - splits every median 2:1 and is the average of the vertex coordinates.
• Circumcentre of a right triangle - mid-point of the hypotenuse, so R = c/2.
• Inradius - r = Area/s; for a right triangle r = (a + b − c)/2.
• Angle at centre - twice the inscribed angle on the same arc; angle in a semicircle is 90°.
• Tangent-radius - perpendicular at the point of contact; equal tangents from an external point.
• Power of a point - PT² = PA·PB links a tangent with any secant from the same point.
• Cyclic quadrilateral - opposite angles supplementary, and that condition is sufficient.
• Rhombus vs rectangle - rhombus diagonals meet at 90° but are unequal; rectangle diagonals are equal but not perpendicular.
• Slope test - parallel if slopes are equal, perpendicular if their product is −1.
• Collinear points - coordinate area = 0 is the fastest check.
• Distance from a point to a line - |ax₁ + by₁ + c|/√(a² + b²); use it for heights and tangency.
• Reflection trick - the shortest path touching a line is a straight line to the mirror image.
• Pythagorean triplets - recognising 3-4-5, 5-12-13, 8-15-17, 7-24-25 and multiples saves the most time in tests.

Formulas and examples

(Lengths in the same unit - cm, m or km; areas in the square of that unit; angles in degrees unless stated.)

Example 1 - a right triangle, completely solved. Legs 9 cm and 12 cm.

• Hypotenuse: c = √(81 + 144) = √225 = 15 cm (a 3-4-5 triplet × 3).
• Area = ½ × 9 × 12 = 54 cm²; s = (9 + 12 + 15)/2 = 18 cm.
• Altitude to the hypotenuse: h = 9 × 12/15 = 7.2 cm (check: ½ × 15 × 7.2 = 54 ✓).
• Hypotenuse segments: 9²/15 = 5.4 cm and 12²/15 = 9.6 cm; they add to 15 ✓.
• Inradius r = Area/s = 54/18 = 3 cm, which matches (a + b − c)/2 = (9 + 12 − 15)/2 = 3 ✓.
• Circumradius R = c/2 = 7.5 cm, the circumcentre being the mid-point of the hypotenuse.

Example 2 - circle theorems. A point P is 13 cm from the centre of a circle of radius 5 cm.

• Tangent length PT = √(13² − 5²) = √144 = 12 cm.
• A secant from P cuts the circle at A (near) and B (far) with PA = 9 cm. Then PT² = PA × PB ⇒ 144 = 9 × PB ⇒ PB = 16 cm, so the chord AB = 16 − 9 = 7 cm.
• A chord of this circle 6 cm long lies at distance √(5² − 3²) = 4 cm from the centre.
• A chord of this circle subtending 70° at the centre subtends 35° at any point on the major arc, and the cyclic quadrilateral formed with that point has the opposite angle 180° − 35° = 145°.

Example 3 - points and the line through them. A(1, 2), B(5, 8).

• AB = √(4² + 6²) = √52 = 2√13 ≈ 7.21 units; mid-point = (3, 5); slope = 6/4 = 1.5.
• Line: y − 2 = 1.5(x − 1) ⇒ 3x − 2y + 1 = 0 (A: 3 − 4 + 1 = 0 ✓; B: 15 − 16 + 1 = 0 ✓).
• The point dividing AB internally in 3 : 1 from A = ((3×5 + 1×1)/4, (3×8 + 1×2)/4) = (4, 6.5).
• Distance of the origin from the line = |0 − 0 + 1|/√(9 + 4) = 1/√13 ≈ 0.277 units.
• Angle with the line x + y = 4 (slope −1): tan θ = |(1.5 + 1)/(1 − 1.5)| = |2.5/(−0.5)| = 5 ⇒ θ = tan⁻¹5 ≈ 78.7°.
• A line perpendicular to AB has slope −2/3; a parallel one keeps slope 1.5.

Example 4 - area, centroid and collinearity. A(1, 2), B(4, 6), C(6, 1).

• Area = ½|1(6 − 1) + 4(1 − 2) + 6(2 − 6)| = ½|5 − 4 − 24| = ½ × 23 = 11.5 square units.
• Centroid = ((1 + 4 + 6)/3, (2 + 6 + 1)/3) = (11/3, 3).
• Are (2, 3), (4, 7) and (7, 13) collinear? Area = ½|2(7 − 13) + 4(13 − 3) + 7(3 − 7)| = ½|−12 + 40 − 28| = 0, so yes - slope check confirms: (7 − 3)/(4 − 2) = 2 and (13 − 7)/(7 − 4) = 2.

Example 5 - reflection, parallel lines and a circle.

• Minimise PA + PB for P on the x-axis with A(1, 2), B(5, 8): reflect A to A′(1, −2); the line A′B has slope (8 + 2)/(5 − 1) = 2.5, so y = 2.5x − 4.5, which meets y = 0 at x = 1.8. P(1.8, 0) and the minimum PA + PB = A′B = √(16 + 100) = √116 = 2√29 ≈ 10.77 units.
• Distance between 3x + 4y − 5 = 0 and 3x + 4y + 15 = 0 = |−5 − 15|/√25 = 4 units; the distance of (3, 4) from the first line = |9 + 16 − 5|/5 = 4 units as well.
• x² + y² − 6x + 4y − 12 = 0 has centre (3, −2) and radius √(9 + 4 + 12) = 5. Since the distance of its centre from 3x + 4y − 5 = 0 is |9 − 8 − 5|/5 = 0.8 < 5, that line is a secant, not a tangent.

Example 6 - polygons. A regular polygon has each exterior angle 24°.

• n = 360/24 = 15 sides; each interior angle = 180 − 24 = 156°; interior angle sum = 13 × 180 = 2,340°; diagonals = 15 × 12/2 = 90.

Where it is used

• Civil engineering and surveying: total-station and chain-survey computations use coordinates, offsets and triangulation; land areas are reported in hectares, acres, guntha or cent in different Indian states, and plot setting-out follows BIS drafting practice (SP 46) and local development-control rules.
• CAD, CAM and manufacturing: AutoCAD, SolidWorks and CNC G-code work entirely in absolute and incremental coordinates, with tangency, fillet radii and bolt-circle geometry computed from these formulas.
• GIS, GPS and logistics: latitude-longitude to projected (UTM) coordinates in QGIS and ArcGIS, geofencing and route distances for delivery and fleet firms (Delhivery, Zomato, Ola, Blinkit).
• Software and game development: collision detection, hit-testing, line-circle intersection and 2-D/3-D transforms in graphics, image processing and robot path planning at ISRO, DRDO, Bosch and Qualcomm India.
• Design and architecture: structural layout grids, truss geometry, staircase and ramp slope (slope = rise/run, capped by NBC accessibility guidance), solar-panel tilt setting.
• Test-prep and teaching roles: aptitude trainers, campus-placement coaches and CBSE/state-board tutors teach exactly this syllabus for Classes 9-12 and competitive exams.

Interview questions

• Q: Why does AAA prove similarity but not congruence? A: Equal angles fix only the shape; the sides can all be scaled by any factor, so the triangles are similar but may differ in size.
• Q: Two similar triangles have areas 49 cm² and 81 cm². What is the ratio of their perimeters? A: The side ratio is √(49/81) = 7 : 9, and perimeters scale linearly, so 7 : 9.
• Q: Where is the circumcentre of a right-angled triangle? A: At the mid-point of the hypotenuse, so the circumradius is half the hypotenuse - the angle in a semicircle being 90° is the same fact.
• Q: A chord 24 cm long lies in a circle of radius 13 cm. How far is it from the centre? A: √(13² − 12²) = 5 cm, using the perpendicular from the centre bisecting the chord.
• Q: How do you check quickly whether three points are collinear? A: Compute the coordinate area ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|; zero means collinear, and equal slopes confirm it.
• Q: What is the condition for two lines to be perpendicular, and what is the exception? A: m₁m₂ = −1; it fails when one line is vertical (undefined slope) and the other horizontal, which are still perpendicular.
• Q: Can a triangle have sides 5, 9 and 14? A: No - 5 + 9 = 14, so the three points are collinear and the "triangle" is degenerate with zero area.
• Q: How do you find the shortest total path from A to B touching a given line? A: Reflect one point across the line and join it to the other; the intersection with the line is the touch point and the straight-line distance is the minimum.
• Q: x² + y² + 2gx + 2fy + c = 0 - when is this not a real circle? A: When g² + f² − c is negative (imaginary circle) or zero (a single point), since the radius is √(g² + f² − c).
• Q: Two circles of radii 8 cm and 3 cm have centres 13 cm apart. How many common tangents? A: 13 > 8 + 3, so the circles are separate and there are four common tangents - two direct, two transverse.
• Q: Why is the centroid used as the centre of gravity in engineering? A: The three medians meet at one point that averages the vertex coordinates; a uniform plate balances on every line through that point, so it is the centre of gravity.
• Q: State the power-of-a-point relation and one use. A: For an external point P, PT² = PA × PB for any secant through P; it lets an unknown chord or tangent length be found from one tangent and one secant measurement.

Common mistakes

• Applying Pythagoras to a triangle that is not right-angled, or treating the longest side as a leg.
• Using the linear ratio for areas of similar figures instead of its square (and the cube for volumes).
• Mixing up the internal and external section formulas - the external one has m − n in the denominator and the signs reversed.
• Reading the centre of x² + y² + 2gx + 2fy + c = 0 as (g, f) rather than (−g, −f), or forgetting the minus sign under the radius root.
• Forgetting the modulus in the area and point-to-line distance formulas and reporting a negative length or area.
• Confusing rhombus and rectangle diagonal properties (perpendicular but unequal versus equal but not perpendicular), and assuming every parallelogram is cyclic - only a rectangle is.
• Dropping the 2 : 1 ratio of the centroid on a median, or assuming the incentre and circumcentre coincide in a non-equilateral triangle.
• Taking a "no solution" from the triangle inequality as an arithmetic slip rather than checking whether the data itself is impossible.

Related topics

Mensuration: areas and perimeters · Mensuration: volumes and surface areas · Algebra: linear and quadratic equations · Squares, cubes, roots and surds · Ratio, proportion and partnership · Data interpretation