Key Concepts
- 1Distance formula
- 2Section formula (internal)
- 3Midpoint formula
- 4Area of triangle (coordinate)
- 5Collinearity condition
- 6Centroid of triangle
- 7Distance from x-axis
- 8Distance from y-axis
- 9State the distance formula between two points (x1, y1) and (x2, y2).
- 10What is the midpoint formula?
Important Formulas & Facts
d = √[(x₂-x₁)² + (y₂-y₁)²]
P = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n))
M = ((x₁+x₂)/2, (y₁+y₂)/2)
½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|
Three points are collinear if the area of the triangle they form = 0.
G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)
|y-coordinate| of the point
|x-coordinate| of the point
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2).
Midpoint of the segment joining (x1, y1) and (x2, y2) is ((x1 + x2)/2, (y1 + y2)/2).
Must-Know Questions
Q1Find the distance between A(3, 4) and B(-1, 2).
d = √[(3-(-1))² + (4-2)²] = √[16 + 4] = √20 = 2√5.
Q2Find the midpoint of the segment joining (4, -6) and (-2, 4).
Midpoint = ((4-2)/2, (-6+4)/2) = (1, -1).
Q3Find the point dividing (1, 3) and (2, 7) in ratio 3:4 internally.
x = (3×2 + 4×1)/7 = 10/7. y = (3×7 + 4×3)/7 = 33/7. Point = (10/7, 33/7).
Q4Find the area of triangle with vertices (2, 3), (5, 7) and (8, 1).
Area = ½|2(7-1) + 5(1-3) + 8(3-7)| = ½|12 - 10 - 32| = ½|-30| = 15 sq units.
Q5Find the ratio in which the y-axis divides the join of (-4, 5) and (3, -7).
On y-axis, x = 0. Let ratio = k:1. 0 = (3k - 4)/(k+1). 3k = 4. k = 4/3. Ratio = 4:3.
Practice Coordinate Geometry
Reinforce what you just revised with practice questions