◆ QA · Geometry

Coordinate Geometry , formulas + CAT PYQs

Focused Geometry kit. The full chapter formula sheet (with explanations & basic examples) is tucked below; every CAT PYQ for Coordinate Geometry is here.

19CAT PYQs
Geometrychapter
Formulas PYQs (19)↩ Full Geometry chapter

Geometry, formula sheet

Show the full Geometry formula sheet (explanations + basic examples)
1Lines & Angles
  • The starting toolkit: most "find the angle" questions are solved just by knowing angles on a line make 180° and walking that around the figure.
  • Angles on a straight line add to 180°; angles around a point add to 360°.
  • Vertically opposite angles are equal.
  • Parallel lines cut by a transversal: corresponding & alternate angles equal; co-interior angles sum to 180°.
  • Exterior angle of a triangle = sum of the two remote interior angles.
  • e.g. Two angles sit on a straight line and one is 110°. The other = 180° − 110° = 70°.
2Triangle, basics & area
  • Pick the area formula that matches what you're given: base+height, all three sides (Heron), or two sides and the angle between them.
  • Angle sum = 180°. Sum of any two sides > the third side.
  • Area = ½ × base × height.
  • Heron: Area = √[s(s−a)(s−b)(s−c)], s = (a+b+c)/2.
  • Area = ½·a·b·sinθ (θ = included angle); Area = r × s (r = inradius); Area = abc/(4R) (R = circumradius).
  • e.g. Sides 3, 4, 5: s = 6, Area = √[6·3·2·1] = √36 = 6 (matches ½·3·4).
Area = √[s(s−a)(s−b)(s−c)] , s = (a+b+c)/2 Area = r·s = abc/(4R)
3Cosine & Sine rules
  • Cosine rule links three sides and one angle (use when you have two sides + included angle, or all three sides); sine rule links sides to opposite angles.
  • Cosine rule: c² = a² + b² − 2ab·cosθ.
  • cosθ = (a² + b² − c²)/(2ab).
  • Sine rule: a/sinA = b/sinB = c/sinC = 2R.
  • e.g. Sides 5 and 8 with a 60° angle between them: third side² = 25 + 64 − 2·5·8·½ = 89 − 40 = 49 ⇒ side = 7.
c² = a² + b² − 2ab·cosθ
4Angle-bisector & medians
  • A bisector splits the far side in the ratio of the two sides it sits between; a median goes to the midpoint, and the centroid cuts it 2:1.
  • Angle bisector divides the opposite side in the ratio of the adjacent sides: BD/DC = AB/AC.
  • Apollonius: b² + c² = 2m² + ½a² (m = median to side a).
  • Median of isosceles (b = c): m² = b² − a²/4.
  • Centroid divides each median in ratio 2 : 1 from the vertex.
  • e.g. AB = 6, AC = 4, bisector meets BC (length 5) at D: BD:DC = 6:4 = 3:2 ⇒ BD = 3, DC = 2.
b² + c² = 2m² + ½a² (Apollonius)
5Pythagoras & altitude relations
  • In a right triangle the squares of the legs add to the square of the hypotenuse; memorising the triplets saves time in the exam.
  • Right triangle: hypotenuse² = base² + height².
  • Altitude to hypotenuse (AD⊥BC, right-angled at A): AD² = BD·DC, AB² = BD·BC, AC² = CD·BC.
  • Acute: AC² = AB² + BC² − 2·BC·BD; Obtuse: AC² = AB² + BC² + 2·BC·BD.
  • Triplets: 3-4-5, 5-12-13, 8-15-17, 7-24-25.
  • e.g. Legs 6 and 8: hypotenuse = √(36 + 64) = √100 = 10 (a scaled 3-4-5).
6Congruence & similarity
  • Congruent = identical; similar = same shape, scaled. The big CAT lever is that areas of similar figures scale as the square of the side ratio.
  • Congruence: SSS, SAS, ASA, AAS, RHS.
  • Similarity: AA, SSS, SAS. Corresponding sides are proportional.
  • Basic Proportionality (Thales): a line ∥ to one side cuts the others in equal ratios.
  • Ratio of areas of similar triangles = (ratio of sides)².
  • e.g. Two similar triangles with sides in ratio 2:3 have areas in ratio 4:9. If the smaller has area 8, the larger = 18.
Area₁ / Area₂ = (side₁ / side₂)²
7Special triangles
  • Equilateral and the two "set-square" triangles (30-60-90, 45-45-90) have fixed side ratios, recognise them and you can write down sides instantly.
  • Equilateral side a: Area = (√3/4)a², height = (√3/2)a, R = a/√3, r = a/(2√3).
  • 30-60-90 sides ratio 1 : √3 : 2.
  • 45-45-90 sides ratio 1 : 1 : √2.
  • From an interior point of an equilateral triangle, sum of ⊥s to the three sides = its height.
  • e.g. Equilateral triangle of side 4: area = (√3/4)·16 = 4√3 and height = (√3/2)·4 = 2√3.
Equilateral Area = (√3/4)·a²
8Geometric centres
  • Four "centres", each the meeting point of a different set of cevians; the incentre and circumcentre are the ones that show up most in area/radius questions.
  • Centroid, intersection of medians (2:1).
  • Incentre, intersection of angle bisectors, centre of inscribed circle.
  • Circumcentre, intersection of ⊥ bisectors of sides, centre of circumscribed circle.
  • Orthocentre, intersection of altitudes.
  • e.g. In a right triangle the circumcentre is the midpoint of the hypotenuse, so a 6-8-10 triangle has circumradius = 10/2 = 5.
9Circle, basics
  • Two workhorses: the angle at the centre is twice the angle at the rim on the same arc, and any angle drawn on a diameter is a right angle.
  • Circumference = 2πr; Area = πr².
  • Equal chords subtend equal angles at the centre & are equidistant from it.
  • ⊥ from the centre bisects the chord.
  • Angle at the centre = 2 × angle at the circumference on the same arc.
  • Angle in a semicircle = 90°.
  • e.g. An arc subtends 40° at the centre, so it subtends 40°/2 = 20° at any point on the major arc.
10Chords, tangents, secants
  • "Power of a point": from any point, the products of the two distances to the circle along a line are equal, chords, secants and tangents all obey it.
  • Two chords meeting at P: PA·PB = PC·PD.
  • Tangent-secant from external P: PA·PB = PT².
  • Tangent ⊥ radius at the point of contact; tangents from an external point are equal.
  • Alternate segment theorem: tangent-chord angle = angle in the alternate segment.
  • e.g. Two chords cross with parts 3 & 8 on one and 4 & x on the other: 3·8 = 4·x ⇒ x = 6.
PA·PB = PC·PD ; PT² = PA·PB
11Cyclic quadrilateral & tangents to 2 circles
  • If all four corners lie on a circle, opposite angles are supplementary; the tangent formulas give the straight-line distance between two circles' touch points.
  • Cyclic quad: opposite angles sum to 180°; exterior angle = opposite interior angle.
  • Ptolemy: AB·CD + BC·DA = AC·BD.
  • A parallelogram inscribed in a circle is a rectangle.
  • Direct common tangent = √[d² − (r₁−r₂)²]; Transverse = √[d² − (r₁+r₂)²].
  • e.g. In a cyclic quad one angle is 70°, so its opposite angle = 180° − 70° = 110°.
Direct tangent = √[d² − (r₁−r₂)²]
12Quadrilaterals
  • Each special quadrilateral has its own area shortcut, base×height for parallelograms, half-product of diagonals for a rhombus, average of parallel sides times height for a trapezium.
  • Parallelogram: opposite sides & angles equal; diagonals bisect each other. Area = base × height.
  • Rectangle: all angles 90°, diagonals equal. Square: all sides equal + 90°.
  • Rhombus: all sides equal; diagonals ⊥ & bisect each other. Area = ½·d₁·d₂.
  • Trapezium: one pair of parallel sides. Area = ½(sum of parallel sides) × height.
  • e.g. Rhombus with diagonals 6 and 8: area = ½·6·8 = 24; trapezium with parallel sides 5 & 9, height 4: area = ½·(5+9)·4 = 28.
Rhombus Area = ½·d₁·d₂ ; Trapezium = ½(a+b)·h
13Polygons
  • Everything flows from "(n−2)·180° of total interior angle"; for a regular polygon the quick route is via the exterior angle, which is just 360°/n.
  • Sum of interior angles = (n − 2)·180°.
  • Each interior angle (regular) = 180° − 360°/n.
  • Each exterior angle (regular) = 360°/n; all exterior angles sum to 360°.
  • Number of diagonals = n(n − 3)/2.
  • e.g. A regular hexagon (n = 6): each exterior angle = 360°/6 = 60°, so each interior angle = 120°; diagonals = 6·3/2 = 9.
Interior sum = (n−2)·180° ; diagonals = n(n−3)/2
14Regular hexagon
  • Think of it as 6 equilateral triangles glued at the centre, that single picture gives its area, diagonals and angles.
  • Side s: Area = (3√3/2)·s².
  • It is 6 equilateral triangles of side s.
  • Longer diagonal = 2s; shorter diagonal = √3·s.
  • Interior angle = 120°.
  • e.g. Hexagon of side 2: area = (3√3/2)·4 = 6√3; long diagonal = 4, short diagonal = 2√3.
Hexagon Area = (3√3/2)·s²
152D mensuration, perimeters & areas
  • The basic flat-shape formulas; a sector is just a fraction θ/360 of the whole circle for both its arc and its area.
  • Square: P = 4a, Area = a², diagonal = a√2.
  • Rectangle: P = 2(l+b), Area = l·b, diagonal = √(l²+b²).
  • Circle: C = 2πr, Area = πr².
  • Sector (angle θ): arc = (θ/360)·2πr, area = (θ/360)·πr².
  • e.g. A 90° sector of a radius-6 circle is ¼ of it: area = ¼·π·36 = 9π, arc = ¼·2π·6 = 3π.
16Cube & cuboid
  • A cube is just a cuboid with l = b = h; the space diagonal (corner-to-corner through the body) uses a 3-term Pythagoras.
  • Cuboid: Volume = l·b·h; TSA = 2(lb + bh + hl); LSA (4 walls) = 2(l+b)h; diagonal = √(l²+b²+h²).
  • Cube edge a: Volume = a³; TSA = 6a²; LSA = 4a²; diagonal = a√3.
  • Sum of all 12 edges: cuboid 4(l+b+h), cube 12a.
  • e.g. Cube of edge 3: volume = 27, TSA = 6·9 = 54, space diagonal = 3√3; a 2×3×6 cuboid has diagonal √(4+9+36) = √49 = 7.
Cuboid V = l·b·h ; TSA = 2(lb+bh+hl)
17Cylinder & cone
  • A cone holds exactly one-third of the cylinder with the same base and height; its slant height is the hypotenuse of the radius-and-height right triangle.
  • Cylinder: Volume = πr²h; CSA = 2πrh; TSA = 2πr(r+h).
  • Cone slant l = √(r²+h²); Volume = ⅓πr²h; CSA = πrl; TSA = πr(r+l).
  • Frustum volume = ⅓πh(R² + r² + Rr).
  • e.g. Cone with r = 3, h = 4: slant = √(9+16) = 5, CSA = π·3·5 = 15π, volume = ⅓·π·9·4 = 12π.
Cone V = ⅓πr²h ; CSA = πrl , l = √(r²+h²)
18Sphere & hemisphere & prism
  • The key exam idea is "recasting": when one solid is melted into another, volume stays the same even though surface area changes.
  • Sphere: Volume = (4/3)πr³; Surface area = 4πr².
  • Hemisphere: Volume = (2/3)πr³; CSA = 2πr²; TSA = 3πr².
  • Prism: Volume = base area × height; LSA = base perimeter × height.
  • Recast objects keep volume constant.
  • e.g. Sphere of radius 3: volume = (4/3)·π·27 = 36π, surface area = 4·π·9 = 36π.
Sphere V = (4/3)πr³ ; SA = 4πr²
19Coordinate geometry, distance & section
  • Distance is just Pythagoras on the coordinate differences; the midpoint is the special case of the section formula with ratio 1:1.
  • Distance = √[(x₂−x₁)² + (y₂−y₁)²].
  • Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2).
  • Section (ratio m:n internal) = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)).
  • Centroid = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3).
  • e.g. Distance from (1, 2) to (4, 6) = √(3² + 4²) = 5; their midpoint = (2.5, 4).
d = √[(x₂−x₁)² + (y₂−y₁)²]
20Coordinate geometry, area & slope
  • Slope = rise over run. Equal slopes mean parallel; slopes multiplying to −1 mean perpendicular. The area formula needs only the three vertices.
  • Area = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|.
  • Slope of line through two points m = (y₂−y₁)/(x₂−x₁).
  • Parallel lines: m₁ = m₂. Perpendicular: m₁·m₂ = −1.
  • e.g. Triangle (0,0), (4,0), (0,3): area = ½|0(0−3)+4(3−0)+0| = ½·12 = 6.
Area = ½|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)|
21Coordinate geometry, lines & circle
  • For the general circle, halve the x- and y-coefficients (with a sign flip) to read off the centre, then back out the radius.
  • Slope-intercept: y = mx + c. Point-slope: y − y₁ = m(x − x₁).
  • ⊥ distance of (x₁,y₁) from ax+by+c=0 = |ax₁+by₁+c|/√(a²+b²).
  • Distance between parallel lines = |c₂−c₁|/√(a²+b²).
  • Circle: (x−h)² + (y−k)² = r²; general x²+y²+2gx+2fy+c=0, centre (−g,−f), r = √(g²+f²−c).
  • e.g. Distance of (0,0) from 3x + 4y − 10 = 0 = |−10|/√(9+16) = 10/5 = 2.
dist = |ax₁+by₁+c| / √(a²+b²)
19 CAT questions

Coordinate Geometry, CAT PYQs

Coordinate Geometry

Coordinate Geometry. Distance, midpoint/section, area by coordinates, slopes and line/circle equations, often blended with pure geometry.
HardCAT 1999

Directions (Q. 1 and 2): Answer the questions based on the following information. A rectangle PRSU, is divided into two smaller rectangles PQTU, and QRST by the line TQ. PQ = 10 cm. QR = 5 cm and RS = 10 cm. Points A, B, F are within rectangle PQTU, and points C, D, E are within the rectangle QRST. The closest pair of points among the pairs (A, C), (A, D), (A, E), (F, C), (F, D), (F, E), (B, C), (B, D), (B, E) are 10√3 cm apart.

Which of the following statements is necessarily true?

PQR STU PQTU (A,B,F) QRST (C,D,E) PQ=10RS=10
  • (1) The closest pair of points among the six given points cannot be (F, C)
  • (2) Distance between A and B is greater than that between F and C.
  • (3) The closest pair of points among the six given points is (C, D), (D, E), or (C, E).
  • (4) None of the above
Show solution
(4) None of the above. We have not been given the distances between any two points, so none of the three statements can be guaranteed.
HardCAT 1999

(Same figure as Q1.) AB > AF > BF and CD > DE > CE, and BF = 6√5 cm. Which of the following is the closest pair of points among all the six given points?

  • (1) B, F
  • (2) C, D
  • (3) A, B
  • (4) None of these
Show solution
(4) None of these. Since CD > DE, option (2) cannot be the answer. Similarly, since AB > AF, option (3) cannot be the answer. We are not sure about the positions of points B and F. Hence, (1) cannot be the answer.
EasyCAT 1999

Directions (Q. 3 and 4): A robot is moved by feeding it with a sequence of instructions. GOTO (x, y), move to the point with coordinates (x, y) no matter where you are currently. WALKX(p), move parallel to the X-axis through a distance of p, in the positive direction if p is positive and in the negative direction if p is negative. WALKY(p), move parallel to the Y-axis through a distance of p, with the same sign convention.

The robot reaches point (6, 6) when a sequence of three instructions is executed, the first of which is a GOTO (x, y) instruction, the second is WALKX(2) and the third is WALKY(4). What are the value of x and y?

  • (1) 2, 4
  • (2) 0, 0
  • (3) 4, 2
  • (4) 2, 2
Show solution
(3) 4, 2. Work backwards from (6, 6). Before WALKY(4) the point was (6, 6−4) = (6, 2); before WALKX(2) it was (6−2, 2) = (4, 2). So GOTO(x, y) = GOTO(4, 2) ⇒ x = 4, y = 2.
EasyCAT 1999

(Same robot as Q3.) The robot is initially at (x, y), x > 0 and y < 0. The minimum number of instructions needed to be executed to bring it to the origin (0, 0) if you are prohibited from using the GOTO instruction is:

  • (1) 2
  • (2) 1
  • (3) x + y
  • (4) 0
Show solution
(1) 2. Two instructions are needed, one parallel to the X-axis and the other parallel to the Y-axis: WALKX(−x) then WALKY(−y).
ModerateCAT 2000

ABCD is a rhombus with the diagonals AC and BD intersecting at the origin on the xy-plane. The equation of the straight line AD is x + y = 1. What is the equation of BC?

  • (1) x + y = −1
  • (2) x − y = −1
  • (3) x + y = 1
  • (4) None of these
Show solution
(1) x + y = −1. AD ∥ BC (opposite sides of rhombus), so BC has the same slope: x + y = k. Diagonals bisect at the origin, so B and C are reflections of A and D ⇒ k = −1. BC: x + y = −1.
EasyCAT 2002

The area of the triangle with the vertices (a, a), (a + 1, a) and (a, a + 2) is:

  • (1) a³
  • (2) 1
  • (3) 0
  • (4) None of these
Show solution
(2) 1. The legs along the axes have lengths 1 and 2, with a right angle at (a, a). Area = ½·1·2 = 1 (independent of a).
ModerateCAT 2005

Four points A, B, C and D lie on a straight line in the XY-plane, such that AB = BC = CD, and the length of AB is 1 metre. An ant at A wants to reach a sugar particle at D. But there are insect repellents kept at points B and C. The ant would not go within one metre of any insect repellent. The minimum distance in metres the ant must traverse to reach the sugar particle is:

  • (1) 3√2
  • (2) 1 + π
  • (3) 4π/3
  • (4) 5
Show solution
(2) 1 + π. The ant detours around each repellent on a quarter-circle of radius 1 and crosses the gap on a straight 1 m. Total = 1 + 2·(π/2) = 1 + π.
ModerateCAT 2017

The shortest distance of the point (½, 1) from the curve y = |x − 1| + |x + 1| is:

  • (1) 1
  • (2) 0
  • (3) √2
  • (4) √(3/2)
Show solution
(1) 1. For |x| ≤ 1 the curve is the horizontal line y = 2; for |x| > 1 it slopes away. The nearest point to (½, 1) lies on y = 2 directly above, giving distance 2 − 1 = 1.
ModerateCAT 2017

The points (2, 5) and (6, 3) are two end points of a diagonal of a rectangle. If the other diagonal has the equation y = 3x + c, then c is:

  • (1) −5
  • (2) −6
  • (3) −7
  • (4) −8
Show solution
(4) −8. Diagonals bisect each other, so the other diagonal passes through the midpoint of the first = (4, 4). Substitute: 4 = 3·4 + c ⇒ c = −8.
ModerateCAT 2018

A triangle ABC has area 32 sq units and its side BC, of length 8 units, lies on the line x = 4. Then the shortest possible distance between A and the point (0, 0) is:

  • (1) 8 units
  • (2) 4 units
  • (3) 2√2 units
  • (4) 4√2 units
Show solution
(2) 4 units. Height from A to BC (line x = 4) = 2·Area/BC = 64/8 = 8, so A's x-coordinate is 4 ± 8 = 12 or −4. To minimise distance to origin, take A = (−4, 0): distance = 4 units.
HardCAT 2019TITA

With rectangular axes of co-ordinates, the number of paths from (1, 1) to (8, 10) via (4, 6), where each step from any point (x, y) is either to (x, y + 1) or to (x + 1, y), is:

Show solution
3920. (1,1)→(4,6): 3 right + 5 up ⇒ ⁸C₃ = 56 paths. (4,6)→(8,10): 4 right + 4 up ⇒ ⁸C₄ = 70 paths. Total = 56·70 = 3920.
ModerateCAT 2019TITA

Let T be the triangle formed by the straight line 3x + 5y − 45 = 0 and the co-ordinate axes. Let the circumcircle of T have radius of length L, measured in the same unit as the coordinate axes. Then, the integer closest to L is:

Show solution
9. Intercepts (15, 0), (0, 9) and origin form a right triangle. Circumradius = ½·hypotenuse = ½·√(15²+9²) = ½·√306 ≈ ½·17.49 = 8.74 ⇒ closest integer 9.
ModerateCAT 2020

The area, in sq. units, enclosed by the lines x = 2, y = |x − 2| + 4, the X-axis and the Y-axis is equal to:

  • (1) 6
  • (2) 8
  • (3) 12
  • (4) 10
Show solution
(4) 10. At x = 2, y = 4 (point A height); at x = 0, y = 6 (OC). It is a trapezium with parallel sides 4 and 6 and width 2. Area = ½·(4+6)·2 = 10.
ModerateCAT 2020

The vertices of a triangle are (0, 0), (4, 0) and (3, 9). The area of the circle passing through these three points is:

  • (1) 205π/9
  • (2) 123π/7
  • (3) 12π/5
  • (4) 14π/3
Show solution
(1) 205π/9. Sides a = PQ = 4, RO = √90, RQ = √82. Area of triangle = ½·4·9 = 18. R = (a·b·c)/(4·Area) ⇒ R² = (82·90)/(4·18)² ·… giving πR² = 205π/9.
ModerateCAT 2020

The points (2, 1) and (−3, −4) are opposite vertices of a parallelogram. If the other two vertices lie on the line x + 9y + c = 0, then c is:

  • (1) 15
  • (2) 12
  • (3) 13
  • (4) 14
Show solution
(4) 14. Diagonals bisect, so the centre = midpoint of (2,1) & (−3,−4) = (−½, −3/2) lies on the line. Substitute: −½ + 9·(−3/2) + c = 0 ⇒ c = 14.
ModerateCAT 2022 · Slot 1

Let ABCD be a parallelogram such that the coordinates of its three vertices A, B, C are (1, 1), (3, 4) and (−2, 8), respectively. Then, the coordinates of the vertex D are:

  • (1) (−4, 5)
  • (2) (−3, 4)
  • (3) (0, 11)
  • (4) (4, 5)
Show solution
(1) (−4, 5). Diagonals bisect: midpoint of AC = midpoint of BD. ((1−2)/2, (1+8)/2) = ((3+x)/2, (4+y)/2) ⇒ x = −4, y = 5. D = (−4, 5).
HardCAT 2023 · Slot 1

Let C be the circle x² + y² + 4x − 6y − 3 = 0 and L be the locus of the point of intersection of a pair of tangents to C with the angle between the two tangents equal to 60°. Then, the point at which L touches the line x = 6 is:

  • (1) (6, 6)
  • (2) (6, 4)
  • (3) (6, 8)
  • (4) (6, 3)
Show solution
(4) (6, 3). Centre (−2, 3), radius = √(4+9+3) = 4. For 60° between tangents, the external point is at distance 8 from centre (since sin30° = 4/CP ⇒ CP = 8). On x = 6: (6+2)² + (y−3)² = 64 ⇒ (y−3)² = 0 ⇒ y = 3. Point (6, 3).
HardCAT 2023 · Slot 2TITA

The area of the quadrilateral bounded by the Y-axis, the line x = 5, and the lines |x − y| − |x − 5| = 2, is:

Show solution
45. For 0 ≤ x ≤ 5, |x−5| = 5−x, so |x−y| = 7−x, giving lines y = 7 and 2x − y − 7 = 0. The trapezium has vertices D(0,7), C(5,7), B(5,3), A(0,−7). Area = ½·(AD + BC)·CD = ½·(14+4)·5 = 45.

CAT 2024 & 2025, recent

HardCAT 2025 · Slot 1

The (x, y) coordinates of vertices P, Q and R of a parallelogram PQRS are (−3, −2), (1, −5) and (9, 1), respectively. If the diagonal SQ intersects the x-axis at (a, 0), then the value of a is

  • (A) 13/4
  • (B) 29/9
  • (C) 10/3
  • (D) 27/7
Show solution
(B) 29/9. Diagonals bisect: midpoint of PR = (3, −0.5) = midpoint of QS → S = (5, 4). Line SQ slope 9/4; set y = 0 → x = 1 + 20/9 = 29/9.