What is the distance formula in coordinate geometry?

The distance between two points (x1, y1) and (x2, y2) is √((x2−x1)² + (y2−y1)²).

What is the section formula?

If a point divides the line joining (x1, y1) and (x2, y2) in the ratio m:n, its coordinates are ((mx2+nx1)/(m+n), (my2+ny1)/(m+n)).

How to find the area of a triangle using coordinates?

Area = ½ |x1(y2−y3) + x2(y3−y1) + x3(y1−y2)| using the coordinates of its vertices.

Are these notes helpful for NEET, SSC, and Railway exams?

Yes, these notes help build strong fundamentals for coordinate geometry, useful for CBSE Class 10 as well as NEET, SSC, Railway, and Police Bharti exams.

Coordinate Geometry — Class 10 Maths Notes

1. Coordinate Plane & Terminology

The coordinate plane has two perpendicular axes: x-axis (horizontal) and y-axis (vertical). A point P is represented by ordered pair (x, y). Quadrants: I (+,+), II (−,+), III (−,−), IV (+,−).

2. Distance Formula

Distance between points P(x₁,y₁) and Q(x₂,y₂) is derived from Pythagoras:

PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]

Example: Distance between (2,3) and (5,7) = √[(3)² + (4)²] = 5.

3. Midpoint Formula

Midpoint M of PQ where P(x₁,y₁), Q(x₂,y₂):

M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )

Useful for bisectors and coordinate geometry proofs.

4. Section Formula (Internal & External)

Point dividing PQ in ratio m:n (internal division) has coordinates:

( (m x₂ + n x₁)/(m + n) , (m y₂ + n y₁)/(m + n) )

For external division, replace (m + n) with (m − n) with sign conventions accordingly.

Example: Point dividing (2,1) and (8,5) in ratio 1:2 internally → ((1×8 + 2×2)/3, (1×5 + 2×1)/3) = (12/3,7/3) = (4,7/3).

5. Slope of a Line

Slope (m) between two points (x₁,y₁) and (x₂,y₂):

m = (y₂ − y₁)/(x₂ − x₁)

Interpretation: rise/run. Vertical lines have undefined slope; horizontal lines have slope 0.

6. Forms of Equation of a Line

Point-slope form: y − y₁ = m(x − x₁)
Slope-intercept form: y = mx + c (c = y-intercept)
Two-point form: (y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁)
Intercept form: x/a + y/b = 1 (a and b are x- and y-intercepts)
General form: Ax + By + C = 0

Find equation of line through (1,2) and (3,6): slope m = (6−2)/(3−1)=2 ⇒ y − 2 = 2(x − 1) ⇒ y = 2x.

7. Area of Triangle using Coordinates

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

Area = (1/2) | x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂) |

Use this to check collinearity (area = 0) or compute area quickly from coordinates.

8. Solved Examples

Example 1. Find midpoint and distance between P(−1,4) and Q(5,−2).

Midpoint = ((−1+5)/2, (4−2)/2) = (2,1). Distance = √[(6)² + (−6)²] = √72 = 6√2.

Example 2. Find equation of line through (2,3) with slope −1/2: y − 3 = −1/2(x − 2) ⇒ y = −1/2 x + 4.

Example 3. Area of triangle with vertices (0,0), (4,0), (4,3) = (1/2)×4×3 = 6. Using formula gives same result.

9. Practice Questions

Find distance between (−3,2) and (4,−1).
Find equation of line passing through (1,1) and (2,3).
Find coordinates of point dividing (0,0) and (9,6) in ratio 2:1 internally.
Find area of triangle with vertices (1,2), (4,6), (6,2).

10. Quick Tips & Exam Strategy

Memorize distance, midpoint and section formulas — they are used repeatedly.
For area problems, prefer determinant formula for accuracy.
Check collinearity using area = 0 formula instead of slope comparisons when coordinates are messy.
Practice converting between different line forms (point-slope, slope-intercept, general).