Coordinate Geometry
Coordinate geometry connects algebra and geometry by placing points on a plane using coordinates (x, y). In Class 10 you learn to find the distance between two points, the coordinates of a point that divides a line segment in a given ratio, the midpoint of a segment, and the area of a triangle from its vertices. The formulas are direct and very scoring.
Learning objectives
- Find the distance between two points using the distance formula.
- Use the section formula to divide a segment in a given ratio.
- Find the midpoint of a line segment.
- Find the area of a triangle from the coordinates of its vertices.
- Apply these formulas to verify shapes and collinearity.
Key concepts
The distance formula
The distance between two points A(x₁, y₁) and B(x₂, y₂) is √[(x₂ − x₁)² + (y₂ − y₁)²]. The distance of a point from the origin is √(x² + y²).
The section formula
The point that divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m : n has coordinates ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)).
Midpoint
The midpoint of A(x₁, y₁) and B(x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2) — the special case of the section formula with ratio 1 : 1.
Area of a triangle
For vertices (x₁, y₁), (x₂, y₂), (x₃, y₃), the area is ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. If the area is zero, the three points are collinear.
Important formulas
Distance
√[(x₂ − x₁)² + (y₂ − y₁)²]
Section (ratio m:n)
((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n))
Midpoint
((x₁ + x₂)/2, (y₁ + y₂)/2)
Area of triangle
½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
Key definitions
- Coordinates
- An ordered pair (x, y) locating a point relative to the x- and y-axes.
- Collinear points
- Three or more points that lie on the same straight line (the triangle's area is zero).
- Origin
- The point (0, 0) where the axes meet.
Solved examples
Q1. Find the distance between A(3, 4) and the origin.
Solution: Distance = √(3² + 4²) = √(9 + 16) = √25 = 5 units.
Q2. Find the midpoint of A(2, 3) and B(4, 7).
Solution: Midpoint = ((2 + 4)/2, (3 + 7)/2) = (3, 5).
Q3. Find the point dividing the segment joining (−1, 7) and (4, −3) in the ratio 2 : 3.
Solution: x = (2·4 + 3·(−1))/(2 + 3) = (8 − 3)/5 = 1; y = (2·(−3) + 3·7)/5 = (−6 + 21)/5 = 3. The point is (1, 3).
Common mistakes to avoid
- Subtracting coordinates in the wrong order inside the squares (the square fixes the sign, but be careful).
- Swapping m and n with the wrong points in the section formula.
- Forgetting the modulus (| |) when finding area, which can give a negative value.
- Mixing up x- and y-coordinates.
Coordinate Geometry — MCQ Quiz
11 questions with instant feedback. Use number keys 1–4 to answer.
The distance of the point (3, 4) from the origin is:
Practice questions
Short answer
Find the distance between (2, 3) and (5, 7).
√[(5−2)² + (7−3)²] = √(9 + 16) = √25 = 5.
What is the midpoint of (1, 1) and (5, 5)?
(3, 3).
Where does a point on the y-axis have its x-coordinate?
Its x-coordinate is 0.
Long answer
Show that the points A(1, 7), B(4, 2), C(−1, −1) and D(−4, 4) are the vertices of a square.
Compute all four sides: AB = BC = CD = DA = √34 (each from the distance formula), and the diagonals AC = BD = √68. Equal sides and equal diagonals mean ABCD is a square.
Find the area of the triangle with vertices (1, 1), (4, 1) and (1, 5).
Area = ½ |1(1 − 5) + 4(5 − 1) + 1(1 − 1)| = ½ |−4 + 16 + 0| = ½ × 12 = 6 square units.
HOTS (Higher Order Thinking)
Find the ratio in which the point (−4, 6) divides the segment joining A(−6, 10) and B(3, −8).
Let the ratio be k : 1. Then −4 = (3k − 6)/(k + 1) ⇒ −4k − 4 = 3k − 6 ⇒ −7k = −2 ⇒ k = 2/7. So the ratio is 2 : 7.
If (3, a) is the midpoint of (1, 4) and (5, 8), find a.
Midpoint y = (4 + 8)/2 = 6, so a = 6 (and the x checks: (1 + 5)/2 = 3).
Quick revision
Revision notes
- Distance: √[(x₂ − x₁)² + (y₂ − y₁)²].
- Section (m:n): ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n)).
- Midpoint: ((x₁ + x₂)/2, (y₁ + y₂)/2).
- Area = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|; zero ⇒ collinear.
Key takeaways
- The midpoint formula is just the section formula at 1 : 1.
- Use the area formula to test collinearity quickly.
- Distance + side comparison proves squares, rhombi and isosceles triangles.
Frequently asked questions
What does coordinate geometry let me do?
It lets you compute lengths, divide segments, find midpoints and areas, and identify shapes using algebra instead of construction.
How do I check whether three points are collinear?
Find the area of the triangle they form; if it is zero, the points are collinear.
Is the distance always positive?
Yes — the square root in the formula always gives a non-negative length.