Coordinate Geometry
CBSE Class 10 · Mathematics · Notes, formulas and practice questions
Class 10 Coordinate Geometry gives you the distance and section formulae to find lengths, divide line segments, check collinearity, and identify quadrilaterals from the coordinates of their vertices.
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This chapter applies the coordinate plane you studied in Class 9 to solve geometric problems algebraically. It introduces two central tools: the distance formula, which uses coordinates to find the length of a segment, and the section formula, which finds the coordinates of a point that cuts a segment in a given ratio. These tools, along with the midpoint formula as a special case, let you answer questions about lines and polygons without plotting points. You will often be asked to check whether points lie on one straight line or to name the type of quadrilateral formed by four points.
The distance formula comes from Pythagoras’ theorem: it measures the hypotenuse of a right triangle formed by the horizontal and vertical differences between two points. Comparing three distances is the standard way to show collinearity. If the sum of the two smaller distances equals the largest distance, the three points must lie on a straight line. Distance also lets you compare sides and diagonals of a quadrilateral, helping you distinguish a square from a rhombus or a rectangle.
The section formula solves the reverse problem as well. If you know the coordinates of a point on a segment, you can put those coordinates into the formula with unknown m and n and solve for the ratio. When the ratio is 1:1 you get the midpoint formula, which is especially useful because each coordinate of the midpoint is the average of the two endpoints. Working with ratios is a common exam skill, and you must identify which endpoint is first and which is second, because the formula is not symmetric.
To classify a quadrilateral, compute the lengths of all four sides and the two diagonals. A quadrilateral with opposite sides equal is a parallelogram; if a parallelogram has equal diagonals it is a rectangle; if all sides are equal it is a rhombus; and a figure that is both a rectangle and a rhombus is a square. An equally fast check for a parallelogram is to show that the diagonals have the same midpoint, using the midpoint formula once for each diagonal. Always arrange the points in the order given before deciding which are sides and which are diagonals.
Key terms
- Distance formula
- Gives the length of the straight segment joining two points (x₁, y₁) and (x₂, y₂). It is obtained from the Pythagorean theorem using the horizontal difference x₂ − x₁ and vertical difference y₂ − y₁.
- Section formula
- Finds the coordinates of a point that lies on the line segment between two given points and divides it in a given positive ratio. Apply it separately to x- and y-coordinates, using AP:PB = m:n.
- Midpoint formula
- Gives the point exactly halfway between two endpoints. Each coordinate of the midpoint is the average of the corresponding coordinates of the endpoints; it is the special case of the section formula when m:n = 1:1.
- Collinear points
- Points that lie on one and the same straight line. Three points are collinear if the largest of the three distances between pairs equals the sum of the other two distances.
- Ratio of division
- The ratio m:n in which a point P divides a segment AB, written AP:PB. To find an unknown ratio, use the section formula with one coordinate and solve for m/n; both coordinates give the same ratio if P truly lies on AB.
- Parallelogram
- A quadrilateral in which both pairs of opposite sides are equal and parallel. In coordinate geometry it is often verified by showing that the diagonals have the same midpoint, because that makes the diagonals bisect each other.
- Rectangle
- A parallelogram with all four angles equal to 90°. From coordinates, you can identify it as a quadrilateral with opposite sides equal and the diagonals also equal.
- Rhombus
- A quadrilateral in which all four sides are equal. It is also a parallelogram; if its diagonals are unequal it is a non-square rhombus, and if they are equal it is a square.
- Square
- A quadrilateral that is both a rectangle and a rhombus: all sides equal and all angles 90°. Verify from coordinates by checking that all four sides are equal and the two diagonals are also equal.
Formula sheet
| What | Formula | Notes |
|---|---|---|
| Distance between two points | d = √((x₂ − x₁)² + (y₂ − y₁)²) | (x₁, y₁) and (x₂, y₂) are the coordinates of the two points. The distance d is always non-negative. |
| Internal division (section formula) | P = ((m x₂ + n x₁)/(m + n), (m y₂ + n y₁)/(m + n)) | P divides the segment from A(x₁, y₁) to B(x₂, y₂) internally in the ratio m:n, so AP:PB = m:n, with m and n positive. |
| Midpoint formula | M = ((x₁ + x₂)/2, (y₁ + y₂)/2) | M is the midpoint of the segment joining (x₁, y₁) and (x₂, y₂). It is the section formula with m:n = 1:1. |
Practice questions with answers
1. Find the distance between A(2, 3) and B(5, 7).
Using the distance formula, d = √((5 − 2)² + (7 − 3)²) = √(3² + 4²) = √(9 + 16) = √25 = 5. So the distance is 5 units.
2. Find the midpoint of the segment joining P(−1, 4) and Q(7, −2).
Midpoint x = (−1 + 7)/2 = 3, and y = (4 + (−2))/2 = 1. Hence the midpoint is (3, 1).
3. Find the coordinates of the point which divides the segment joining A(1, 2) and B(6, 7) internally in the ratio 2:3.
Let AP:PB = 2:3. By the section formula, x = (2×6 + 3×1)/(2+3) = 15/5 = 3 and y = (2×7 + 3×2)/(2+3) = 20/5 = 4. The required point is (3, 4).
4. Find the ratio in which the point P(4, 6) divides the segment joining A(2, 4) and B(5, 7).
Let AP:PB = m:n. From the x-coordinate, 4 = (5m + 2n)/(m + n) gives m = 2n. The y-coordinate gives the same result: 6 = (7m + 4n)/(m + n) also gives m = 2n. Therefore m:n = 2:1.
5. Show that A(1, 2), B(3, 4) and C(5, 6) are collinear.
AB = √((3−1)² + (4−2)²) = √8 = 2√2, BC = √((5−3)² + (6−4)²) = 2√2, and AC = √((5−1)² + (6−2)²) = √32 = 4√2. Since AB + BC = 2√2 + 2√2 = 4√2 = AC, the largest distance equals the sum of the other two, so the points are collinear.
6. Find k if the points A(k, 0), B(1, 2) and C(3, 4) are collinear.
Let B divide AC internally in the ratio m:n. From the y-coordinate of B, 2 = (4m + 0n)/(m + n) = 4m/(m + n), so m + n = 2m, which gives m = n. Hence B is the midpoint of AC. Using the midpoint formula for the x-coordinate, 1 = (k + 3)/2, so k = −1.
7. Prove that A(1, 2), B(4, 5), C(6, 3) and D(3, 0) are the vertices of a parallelogram.
Find the midpoints of the diagonals. Midpoint of AC is ((1+6)/2, (2+3)/2) = (3.5, 2.5). Midpoint of BD is ((4+3)/2, (5+0)/2) = (3.5, 2.5). Since the diagonals have the same midpoint, they bisect each other, so ABCD is a parallelogram.
8. Show that A(0, 0), B(2, 3), C(4, 0) and D(2, −3) form a rhombus.
Using the distance formula, AB = √((2−0)² + (3−0)²) = √13, BC = √((4−2)² + (0−3)²) = √13, CD = √((2−4)² + (−3−0)²) = √13, and DA = √((0−2)² + (0+3)²) = √13. All four sides are equal, so the quadrilateral is a rhombus.
9. Find the point on the y-axis that is equidistant from A(1, 2) and B(3, 4).
Let the point be P(0, y). Then PA² = (0−1)² + (y−2)² = y² − 4y + 5 and PB² = (0−3)² + (y−4)² = y² − 8y + 25. Equate: y² − 4y + 5 = y² − 8y + 25, giving 4y = 20, so y = 5. Hence the point is (0, 5).
10. If A(2, 3), B(8, 9) and P(x, 7) lies on AB, find x and the ratio in which P divides AB.
Let AP:PB = m:n. Using the y-coordinate of P, 7 = (9m + 3n)/(m + n) gives 7m + 7n = 9m + 3n, so 4n = 2m and m:n = 2:1. Now use the x-coordinate: x = (8×2 + 2×1)/(2+1) = 18/3 = 6. So P divides AB in the ratio 2:1 and x = 6.
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