Coordinate Geometry (also called Analytic Geometry) uses algebraic methods to study geometric properties of figures. We place figures on a coordinate plane with x-axis and y-axis. Every point is identified by an ordered pair (x, y), and we can calculate distances, midpoints, and areas using formulas derived from these coordinates.
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Key Concepts
Distance Formula: The distance between two points A(x1, y1) and B(x2, y2) is:
AB = √((x2 - x1)2 + (y2 - y1)2)
Section Formula: The point P(x, y) that divides the line segment joining A(x1, y1) and B(x2, y2) in the ratio m:n (internally) is:
x = (m · x2 + n · x1) / (m + n)
y = (m · y2 + n · y1) / (m + n)
Midpoint Formula (special case of section formula with m = n = 1):
x = (x1 + x2) / 2
y = (y1 + y2) / 2
Area of a Triangle: For vertices A(x1, y1), B(x2, y2), C(x3, y3):
Area = (1/2) x |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
Collinear Points: Three points are collinear if the area of the triangle formed by them is 0.
Distance from Origin: Distance of point P(x, y) from origin O(0, 0) = √(x2 + y2).
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Worked Examples
Find the distance between A(3, 4) and B(7, 1).
AB = √((7-3)2 + (1-4)2) = √(16 + 9) = √(25) = 5 units.
Find the midpoint of the segment joining P(-1, 4) and Q(3, -2).
Midpoint = ((-1+3)/2, (4+(-2))/2) = (2/2, 2/2) = (1, 1).
Find the point dividing A(1, 2) and B(7, 8) in ratio 1:2 internally.
x = (1 x 7 + 2 x 1) / (1 + 2) = (7 + 2)/3 = 9/3 = 3.
y = (1 x 8 + 2 x 2) / (1 + 2) = (8 + 4)/3 = 12/3 = 4.
Point P = (3, 4).
Find the area of the triangle with vertices A(2, 3), B(4, -1), C(-1, 2).
Area = (1/2)|2(-1-2) + 4(2-3) + (-1)(3-(-1))|
= (1/2)|2(-3) + 4(-1) + (-1)(4)|
= (1/2)|-6 - 4 - 4|
= (1/2) x 14 = 7 sq units.
Show that A(3, 5), B(-1, 2), C(7, 8) are collinear.
Area = (1/2)|3(2-8) + (-1)(8-5) + 7(5-2)|
= (1/2)|3(-6) + (-1)(3) + 7(3)|
= (1/2)|-18 - 3 + 21|
= (1/2)|0| = 0. The points are collinear.
Find the centroid of the triangle with vertices A(4, -2), B(-2, 4), C(5, 5).
Centroid G = ((4 + (-2) + 5)/3, (-2 + 4 + 5)/3) = (7/3, 7/3).
G = (7/3, 7/3).
Find the ratio in which the line segment joining A(-3, 10) and B(6, -8) is divided by the point P(-1, 6).
Let ratio = k:1. -1 = (k x 6 + 1 x (-3)) / (k + 1) = (6k - 3) / (k + 1).
-1(k+1) = 6k - 3. -k - 1 = 6k - 3. 2 = 7k. k = 2/7.
Ratio = 2:7. Verify with y: (2x(-8) + 7x10)/9 = (-16+70)/9 = 54/9 = 6. Correct!
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Key Formulas
- Distance: √((x2-x1)2 + (y2-y1)2)
- Section (internal): ((m · x2+n · x1)/(m+n), (m · y2+n · y1)/(m+n))
- Midpoint: ((x1+x2)/2, (y1+y2)/2)
- Area of triangle: (1/2)|x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|
- Centroid: ((x1+x2+x3)/3, (y1+y2+y3)/3)
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Common mistakes
- Forgetting the absolute value in the area formula — area is always positive.
- Applying the section formula backwards: in m:n dividing AB, the formula uses m times B's coordinates plus n times A's coordinates (not the other way around).
- Mixing up midpoint formula as (x2 - x1)/2 instead of (x1 + x2)/2.
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Summary
Coordinate Geometry transforms geometric problems into algebraic ones. The three main tools — distance formula, section formula, and area formula — are sufficient to solve the vast majority of Class 10 problems. Always double-check coordinate signs and use the modulus for area.