The Coordinate Plane
A point in the plane is located by an ordered pair (x, y):
- x = horizontal distance from origin (positive rightward)
- y = vertical distance from origin (positive upward)
Distance Formula
Distance between A(x₁, y₁) and B(x₂, y₂):
$$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
| Derivation: Draw a right triangle with AB as hypotenuse. Horizontal leg = | x₂−x₁ | , vertical leg = | y₂−y₁ | . By Pythagoras: AB² = (x₂−x₁)² + (y₂−y₁)² |
|---|
Special cases:
- Distance from origin to (x, y) = √(x² + y²)
- Horizontal distance (same y): |x₂ − x₁|
- Vertical distance (same x): |y₂ − y₁|
Worked examples:
Q1. Find distance between A(3, 4) and B(−1, 1). AB = √[(−1−3)² + (1−4)²] = √[16 + 9] = √25 = 5 units
Q2. Show that A(1,2), B(5,4), C(3,8), D(−1,6) form a square. AB = √(16+4) = √20; BC = √(4+16) = √20; CD = √(16+4) = √20; DA = √(4+16) = √20 All sides equal. Check diagonals: AC = √(4+36) = √40; BD = √(36+4) = √40. Equal diagonals → square ✓
Section Formula
Point P(x, y) dividing AB internally in ratio m:n:
$$x = \frac{mx_2 + nx_1}{m+n}, \quad y = \frac{my_2 + ny_1}{m+n}$$
Midpoint formula (m:n = 1:1): $$M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$$
Worked examples:
Q1. Find the point that divides A(4, −3) and B(8, 5) in ratio 3:1. x = (3×8 + 1×4)/(3+1) = (24+4)/4 = 28/4 = 7 y = (3×5 + 1×(−3))/4 = (15−3)/4 = 12/4 = 3 Point = (7, 3)
Q2. Find the ratio in which P(−2, 3) divides A(−3, 5) and B(4, −9). Let ratio = k:1. x: −2 = (4k + (−3))/(k+1) → −2k − 2 = 4k − 3 → 1 = 6k → k = 1/6 Ratio = 1:6
Area of Triangle
For triangle with vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃):
| $$\text{Area} = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | $$ |
|---|
Take the absolute value — area is always positive.
If area = 0, the three points are collinear (lie on the same line).
Worked examples:
Q1. Find area of △ with vertices A(2, 3), B(−1, 0), C(4, −3).
| Area = ½ | 2(0−(−3)) + (−1)(−3−3) + 4(3−0) | |
|---|---|---|
| = ½ | 2(3) + (−1)(−6) + 4(3) | |
| = ½ | 6 + 6 + 12 | = ½ × 24 = 12 sq. units |
Q2. Check if A(1, 2), B(3, 6), C(5, 10) are collinear.
| Area = ½ | 1(6−10) + 3(10−2) + 5(2−6) | |||
|---|---|---|---|---|
| = ½ | 1(−4) + 3(8) + 5(−4) | |||
| = ½ | −4 + 24 − 20 | = ½ | 0 | = 0 → Points are collinear ✓ |
Centroid of a Triangle
The centroid (intersection of medians) of triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃):
$$G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$$
Board Exam Tips
- 5–6 marks from this chapter
- Distance formula is straightforward — careful with signs when squaring differences
- Section formula: write m:n clearly, identify which is m and which is n (ratio in which the first-named point divides)
- Area formula: use the cyclic pattern — x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂). Never forget the ½
- Collinearity is tested by setting area = 0 — very common 2-mark question
- "Find ratio in which P divides AB" — use section formula with k:1, solve for k
- "Find value of k if points are collinear" — set area = 0, solve for k
📚 RD Sharma Mathematics Class 10 — Coordinate Geometry Exercises
📝 Oswaal CBSE Maths Question Bank — Coordinate Geometry