Quadrilaterals and Coordinate Geometry
Quadrilaterals and Coordinate Geometry
Quadrilateral questions test the special properties of the parallelogram family: which have equal diagonals, which have perpendicular ones, which bisect angles. Coordinate geometry puts figures on a grid: distance, midpoint, section and centroid formulas, slopes, and areas from coordinates.
1. Quadrilateral Properties
| Figure | Diagonals bisect each other | Diagonals equal | Diagonals perpendicular | Other |
|---|---|---|---|---|
| Parallelogram | Yes | No | No | Opposite angles equal; adjacent angles add to 180° |
| Rectangle | Yes | Yes | No | All angles 90° |
| Rhombus | Yes | No | Yes | All sides equal; diagonals bisect the angles |
| Square | Yes | Yes | Yes | All of the above |
| Kite | One bisects the other | No | Yes | Two pairs of equal adjacent sides |
Angle sum of a quadrilateral = 360°
Mid-points of the sides of any quadrilateral form a parallelogram
2. Coordinate Formulas
| Formula | For points (x₁, y₁), (x₂, y₂), (x₃, y₃) | ||
|---|---|---|---|
| Distance | √((x₂ − x₁)² + (y₂ − y₁)²) | ||
| Midpoint | ((x₁ + x₂)/2, (y₁ + y₂)/2) | ||
| Section, ratio m : n internally | ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) | ||
| Centroid | ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3) | ||
| Slope | (y₂ − y₁)/(x₂ − x₁) | ||
| Area of triangle | (1/2) × | x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂) |
Line ax + by = c meets the axes at (c/a, 0) and (0, c/b)
Parallel lines: equal slopes. Perpendicular lines: product of slopes = −1
Distance of ax + by + c = 0 from the origin = |c|/√(a² + b²)
Collinear points: area of the triangle = 0 (or equal slopes)
| Quadrant | Sign of (x, y) |
|---|---|
| I | (+, +) |
| II | (−, +) |
| III | (−, −) |
| IV | (+, −) |
✗ The point dividing A and B in ratio 2 : 3 is (2x₁ + 3x₂)/5 | ✓ The ratio number attaches to the far point: (2x₂ + 3x₁)/5
✗ Rhombus diagonals are equal | ✓ Only rectangle and square diagonals are equal; rhombus diagonals are perpendicular
हिंदी नोट: समांतर चतुर्भुज के विकर्ण एक-दूसरे को समद्विभाजित करते हैं; आयत में विकर्ण बराबर और समचतुर्भुज में लंबवत होते हैं। निर्देशांक ज्यामिति में दूरी, मध्य बिंदु, विभाजन और केन्द्रक सूत्र याद रखिए।
Exam Pointer: These patterns had no verified NTPC shift in our check and are tagged syllabus-based. Quadrilateral properties also appear as statement questions, and coordinate geometry is listed in the NTPC Mathematics syllabus; both turn up in CBT-2 more often than CBT-1.
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Angles of parallelograms and quadrilaterals
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. The angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. Find them.
18 parts = 360°, so one part = 20° and the angles are 60°, 80°, 100° and 120°.
Answer: 60°, 80°, 100°, 120°
EXAMATLAS LEVEL
Q. In parallelogram ABCD, ∠A = 3∠B. In rectangle PQRS the diagonals meet at O with ∠POQ = 110°. Find ∠A, and ∠OPQ and ∠OQR.
Adjacent angles add to 180°, so 4∠B = 180°, ∠B = 45° and ∠A = 135°. In the rectangle OP = OQ, so ∠OPQ = ∠OQP = (180° − 110°)/2 = 35°, and ∠OQR = 90° − 35° = 55°.
Answer: 135°; 35° and 55°
Pattern 2: Distance, midpoint and centroid
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find the distance between (2, 3) and (8, 11), and their midpoint.
√(36 + 64) = 10. Midpoint = (5, 7).
Answer: 10; (5, 7)
EXAMATLAS LEVEL
Q. Find the centroid of the triangle with vertices (1, 2), (4, 6) and (7, 1), and check whether the triangle is isosceles.
Centroid = (12/3, 9/3) = (4, 3). Side lengths: from (1, 2) to (4, 6) is 5; from (4, 6) to (7, 1) is √34; from (1, 2) to (7, 1) is √37. No two are equal, so it is scalene.
Answer: (4, 3); not isosceles
Pattern 3: Section formula
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find the point dividing the join of (−2, 3) and (8, −7) in the ratio 2 : 3 internally.
x = (2 × 8 + 3 × (−2))/5 = 2 and y = (2 × (−7) + 3 × 3)/5 = −1.
Answer: (2, −1)
EXAMATLAS LEVEL
Q. In what ratio does the y-axis divide the join of (−3, 4) and (6, −2), and at which point?
On the y-axis x = 0: (6m − 3n)/(m + n) = 0 gives m : n = 1 : 2. Then y = (1 × (−2) + 2 × 4)/3 = 2.
Answer: 1 : 2 at (0, 2)
Pattern 4: Lines, intercepts, area and collinearity
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find the area of the triangle formed by the line 3x + 4y = 24 and the axes.
Intercepts are 8 and 6, so area = (1/2) × 8 × 6 = 24 square units.
Answer: 24 square units
EXAMATLAS LEVEL
Q. For what k are (1, 2), (3, k) and (7, 14) collinear? Also find the area of the triangle with vertices (1, 1), (4, 5) and (7, 2).
Slope from (1, 2) to (7, 14) is 12/6 = 2, so k = 2 + 2 × 2 = 6. Area = (1/2)|1(5 − 2) + 4(2 − 1) + 7(1 − 5)| = (1/2)|3 + 4 − 28| = 10.5 square units.
Answer: k = 6; 10.5 square units
60-Second Revision
- Parallelogram: adjacent angles add to 180°; rectangle diagonals equal; rhombus diagonals perpendicular.
- Distance, midpoint, centroid: learn the three formulas cold.
- Section m : n: the m goes with the second point.
- Area of a triangle from coordinates; area 0 means collinear.
Next Step: Geometry done. Draw a quick figure for every question in the ExamAtlas RRB NTPC 2026 mock tests; most geometry answers come from one theorem once the figure is marked.