📐 Coordinate Geometry — Class 10

Distance formula, section formula, area of triangle using coordinates

1. Distance Formula

📖 Distance Between Two Points

The distance between two points P(x₁, y₁) and Q(x₂, y₂) is:

PQ = √[(x₂ – x₁)² + (y₂ – y₁)²]

This is derived from the Pythagorean theorem where PQ is the hypotenuse.

📖 Distance from Origin

The distance of a point P(x, y) from the origin O(0, 0):

OP = √(x² + y²)

💡 Example 1: Find distance between A(3, 4) and B(7, 1)

AB = √[(7–3)² + (1–4)²] = √[4² + (–3)²] = √[16 + 9] = √25 = 5 units

💡 Example 2: Show that (3, 0), (6, 4), (–1, 3) form a right triangle

Let A(3,0), B(6,4), C(–1,3)

AB = √[(6–3)² + (4–0)²] = √[9+16] = √25 = 5

BC = √[(–1–6)² + (3–4)²] = √[49+1] = √50 = 5√2

CA = √[(3–(–1))² + (0–3)²] = √[16+9] = √25 = 5

AB² + CA² = 25 + 25 = 50 = BC² ✓

Answer: Triangle ABC is right-angled at A (AB = CA → also isosceles right triangle)

💡 Example 3: Find value of y if A(3, y) is equidistant from P(6, 5) and Q(0, –3)

AP = AQ (given)

√[(6–3)² + (5–y)²] = √[(0–3)² + (–3–y)²]

Squaring: 9 + (5–y)² = 9 + (–3–y)²

(5–y)² = (–3–y)² → 25 – 10y + y² = 9 + 6y + y²

16 = 16y → y = 1

2. Section Formula

📖 Internal Division

If point P divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in ratio m:n, then:

P = [(mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n)]

📖 Midpoint Formula

Midpoint M of line segment joining A(x₁, y₁) and B(x₂, y₂):

M = [(x₁ + x₂)/2, (y₁ + y₂)/2]

(This is just section formula with m = n = 1)

💡 Example 1: Section Formula

Q: Find the point which divides the line segment joining A(2, –3) and B(5, 6) in ratio 1:2 internally.

m = 1, n = 2, x₁ = 2, y₁ = –3, x₂ = 5, y₂ = 6

x = (1×5 + 2×2)/(1+2) = (5+4)/3 = 9/3 = 3

y = (1×6 + 2×(–3))/(1+2) = (6–6)/3 = 0/3 = 0

Answer: Point P = (3, 0)

💡 Example 2: Midpoint

Q: Find midpoint of A(–2, 3) and B(4, –1).

M = [(–2+4)/2, (3–1)/2] = [2/2, 2/2] = (1, 1)

💡 Example 3: Finding ratio

Q: In what ratio does point P(4, 5) divide the line joining A(1, 2) and B(7, 8)?

Let ratio = k:1

x-coordinate: (7k + 1)/(k+1) = 4 → 7k+1 = 4k+4 → 3k = 3 → k = 1

Answer: P divides AB in ratio 1:1 (midpoint!)

3. Area of a Triangle

📖 Area Using Coordinates

Area of triangle with vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃):

Area = ½ |x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|

The modulus | | ensures area is always positive.

⚡ Collinearity Condition

Three points A, B, C are collinear (lie on a straight line) if and only if the area of △ABC = 0.

i.e., x₁(y₂–y₃) + x₂(y₃–y₁) + x₃(y₁–y₂) = 0

💡 Example: Find area of triangle

Q: Find area of triangle with vertices A(5, 2), B(4, 7), C(7, –4).

Area = ½ |5(7–(–4)) + 4(–4–2) + 7(2–7)|

= ½ |5(11) + 4(–6) + 7(–5)|

= ½ |55 – 24 – 35|

= ½ |–4| = ½ × 4 = 2 sq. units

💡 Example: Check Collinearity

Q: Are A(2, 3), B(4, 5), C(6, 7) collinear?

Area = ½ |2(5–7) + 4(7–3) + 6(3–5)|

= ½ |2(–2) + 4(4) + 6(–2)| = ½ |–4 + 16 – 12| = ½ |0| = 0

Yes, the points are collinear.

4. Key Points to Remember

Distance formula works for any two points in coordinate plane.
Section formula: internal division gives a point between A and B.
Midpoint is a special case of section formula (m = n).
Area of triangle = 0 means the three points are collinear.
Always take modulus when computing area to get positive value.

📋 Formula Sheet — Coordinate Geometry

Distance Formula

d = √[(x₂–x₁)² + (y₂–y₁)²]

Section Formula (m:n)

x = (mx₂+nx₁)/(m+n)

y = (my₂+ny₁)/(m+n)

Midpoint

M = ((x₁+x₂)/2, (y₁+y₂)/2)

Area of Triangle

½|x₁(y₂–y₃)+x₂(y₃–y₁)+x₃(y₁–y₂)|