Geometric Transformations in JEE Main Math

Geometric transformations — reflection, rotation, translation — appear in JEE Main through "find the image of a point/line/curve" questions in Coordinate Geometry. These are rapid calculation questions if you know the formulas; they become lengthy derivations if you do not. This guide provides the key transformation formulas and shows how JEE applies them.

Reflection (Mirror Image)

Image of point P(x₁, y₁) in the line ax + by + c = 0: the image P'(h, k) satisfies (h − x₁)/a = (k − y₁)/b = −2(ax₁ + by₁ + c)/(a² + b²). This gives: h = x₁ − 2a(ax₁ + by₁ + c)/(a² + b²) and k = y₁ − 2b(ax₁ + by₁ + c)/(a² + b²). Special cases: reflection in x-axis (y = 0): image is (x₁, −y₁). Reflection in y-axis (x = 0): image is (−x₁, y₁). Reflection in y = x: image is (y₁, x₁). Reflection in y = −x: image is (−y₁, −x₁). Reflection in origin: image is (−x₁, −y₁). Reflection in x = a: image is (2a − x₁, y₁). Reflection in y = b: image is (x₁, 2b − y₁).

JEE question: "Find the image of (2, 3) in the line 3x − 4y + 5 = 0." Using the formula: 3(2) + (−4)(3) + 5 = 6 − 12 + 5 = −1. a = 3, b = −4, a² + b² = 25. h = 2 − 2(3)(−1)/25 = 2 + 6/25 = 56/25. k = 3 − 2(−4)(−1)/25 = 3 − 8/25 = 67/25. Image: (56/25, 67/25). Take a free coordinate geometry mock. See our straight lines guide.

Rotation About the Origin

Rotation of point (x, y) by angle θ anticlockwise about the origin: new coordinates (x', y') = (x cosθ − y sinθ, x sinθ + y cosθ). In matrix form: [x'; y'] = [[cosθ, −sinθ]; [sinθ, cosθ]] × [x; y]. Special angles: 90° anticlockwise: (x, y) → (−y, x). 90° clockwise: (x, y) → (y, −x). 180°: (x, y) → (−x, −y). JEE question: "A line has slope 1 (at 45° to x-axis). After rotating the line by 30° anticlockwise, find the new slope." The new angle with x-axis = 45° + 30° = 75°. New slope = tan 75° = tan(45° + 30°) = (1 + tan30°)/(1 − tan30°) = (1 + 1/√3)/(1 − 1/√3) = (√3 + 1)/(√3 − 1) = 2 + √3.

Translation and Shifting

Translating the origin to (h, k): substitute x → X + h, y → Y + k (where X, Y are the new coordinates relative to the new origin). This eliminates first-degree terms in the equation of a conic, converting general form to standard form. Example: x² + y² − 4x + 6y − 12 = 0. Complete the square: (x−2)² + (y+3)² = 25. This is a translation to (h, k) = (2, −3): in new coordinates X = x−2, Y = y+3, the circle is X² + Y² = 25. JEE uses translation to reduce the general second-degree equation to its canonical form (centre/vertex at origin).

Image of Curves Under Transformations

To find the image of a curve f(x, y) = 0 under a transformation: substitute the inverse transformation. For reflection in y = x: replace x with y and y with x in the equation. For rotation by 90°: replace (x, y) with (y, −x) in the equation (since the inverse of a 90° anticlockwise rotation is a 90° clockwise rotation). Example: image of parabola y = x² under reflection in y = x: replace x → y, y → x: x = y² → y² = x (horizontal parabola). This is the inverse function relationship. For the image of a line y = mx + c under reflection in another line: use the reflection formula twice (find images of two points, then get the line through them). For reflection and transformation applied to conic sections see our parabola guide and our coordinate geometry guide.