Class 10 Maths Chapter 7 – Coordinate Geometry | NCERT Solutions, Formulas & CBSE Board Questions (2025–26)

 Class 10 CBSE Maths – Chapter 7: Coordinate Geometry

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🔹 Key Concepts and Formulas

✅ Coordinate Plane

• A plane with two perpendicular lines (axes): x-axis (horizontal) and y-axis (vertical)

• The point of intersection is called the origin (0, 0)

• Coordinates are written as (x, y)

✅ Quadrants

• I Quadrant: (+x, +y)

• II Quadrant: (–x, +y)

• III Quadrant: (–x, –y)

• IV Quadrant: (+x, –y)

✅ Distance Formula

To find distance between two points $A(x_1, y_1)$ and $B(x_2, y_2)$:

$$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

✅ Section Formula

To find coordinates of a point P which divides the line segment AB in the ratio $m:n$:

$$P = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)$$

✅ Midpoint Formula

Special case of section formula when m = n:

$$\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$$

✅ Area of a Triangle (Coordinate Geometry)

Given 3 points $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$:

$$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

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📘 NCERT Solved Examples (Highlights)

Example 1:

Find the distance between points (3, 4) and (7, 1):

$$AB = \sqrt{(7 - 3)^2 + (1 - 4)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$

Example 2:

Find the coordinates of the point which divides the line joining (2, 3) and (4, 5) in the ratio 1:2:

$$P = \left( \frac{1 \times 4 + 2 \times 2}{1 + 2}, \frac{1 \times 5 + 2 \times 3}{1 + 2} \right) = (\frac{8}{3}, \frac{11}{3})$$

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📄 Exercise 7.1 – Distance Formula

Q1. Find the distance between points (2, 3) and (4, 1):

$$= \sqrt{(4 - 2)^2 + (1 - 3)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$$

Q2. Find the distance between (0, 0) and (36, 15):

$$= \sqrt{36^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39$$

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📄 Exercise 7.2 – Section Formula

Q1. Find coordinates of point dividing the segment joining (1, -2) and (3, 6) in the ratio 2:1:

$$P = \left( \frac{2 \times 3 + 1 \times 1}{3}, \frac{2 \times 6 + 1 \times (-2)}{3} \right) = \left( \frac{7}{3}, \frac{10}{3} \right)$$

Q2. Find coordinates of the midpoint of line joining (4, -1) and (-2, 3):

$$= \left( \frac{4 + (-2)}{2}, \frac{-1 + 3}{2} \right) = (1, 1)$$

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📄 Exercise 7.3 – Area of Triangle

Q1. Find the area of triangle with vertices (2, 3), (4, 5), (6, 7):

$$= \frac{1}{2} |2(5 - 7) + 4(7 - 3) + 6(3 - 5)| = \frac{1}{2} |-4 + 16 -12| = \frac{1}{2} \times 0 = 0$$

⇒ Points are collinear

Q2. Find the area of triangle with vertices A(1, 1), B(4, 4), C(1, 6):

$$= \frac{1}{2} |1(4 - 6) + 4(6 - 1) + 1(1 - 4)| = \frac{1}{2} |-2 + 20 - 3| = \frac{1}{2} \times 15 = 7.5$$

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🔁 Summary and Revision Notes

• Distance Formula helps find the length of a segment between two points

• Section Formula finds coordinates dividing a segment in a ratio

• Midpoint is a special case of section formula

• Area Formula useful for checking collinearity of points (area = 0 ⇒ collinear)

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📘 CBSE Previous Year Questions – Coordinate Geometry

✅ 1-Mark

(2020) What is the distance between point A(3, -4) and origin?

$$= \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

✅ 2-Mark

(2021) Find the coordinates of the point which divides the line joining A(1, 2) and B(3, 4) in the ratio 1:1.

$$Midpoint = \left( \frac{1+3}{2}, \frac{2+4}{2} \right) = (2, 3)$$

✅ 3-Mark

(2018) Find the area of the triangle with vertices (2, 3), (4, -1), (-1, 2):

$$= \frac{1}{2} |2(-1 - 2) + 4(2 - 3) + (-1)(3 + 1)| = \frac{1}{2} |-6 -4 -4| = \frac{1}{2} \times 14 = 7$$

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