Coordinate Geometry — Class 9 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Coordinate Geometry" — 8 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Coordinate Geometry" — 8 important questions with detailed answers for CBSE board e…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

Plot the following points on a coordinate plane and determine which quadrant …
Find the distance between points P(1, 3) and Q(4, 7).
Find the coordinates of the midpoint of the line segment joining A(-2, 4) and…
Show that the points A(0, 0), B(3, 4), and C(6, 8) are collinear.
Find the area of a triangle with vertices at A(1, 2), B(4, 2), and C(4, 5).
A point lies on the x-axis. If its distance from (2, 3) is 5 units, find the …
+ 2 more questions in the full chapter

Solutions Summary:

Question	Status
Plot the following points on a coordinate plane and deter…	✓ Solved
Find the distance between points P(1, 3) and Q(4, 7).	✓ Solved
Find the coordinates of the midpoint of the line segment …	✓ Solved
Show that the points A(0, 0), B(3, 4), and C(6, 8) are co…	✓ Solved
Find the area of a triangle with vertices at A(1, 2), B(4…	✓ Solved
A point lies on the x-axis. If its distance from (2, 3) i…	✓ Solved

Showing 6 of 8 questions

Q1: Plot the following points on a coordinate plane and determine which quadrant each belongs to: A(3, 4), B(-2, 5), C(-3, -2), D(4, -1).

Step 1: Understand quadrants Quadrant I: x > 0, y > 0 Quadrant II: x < 0, y > 0 Quadrant III: x < 0, y < 0 Quadrant IV: x > 0, y < 0 Step 2: Classify each point Point A(3, 4): x = 3 > 0, y = 4 > 0 → Quadrant I Point B(-2, 5): x = -2 < 0, y = 5 > 0 → Quadrant II Point C(-3, -2): x = -3 < 0, y = -2 < 0 → Quadrant III Point D(4, -1): x = 4 > 0, y = -1 < 0 → Quadrant IV Step 3: Plot on coordinate plane - Mark origin O at (0, 0) - A is 3 units right,...

Q2: Find the distance between points P(1, 3) and Q(4, 7).

Points: P(1, 3) and Q(4, 7) Use distance formula: d = √[(x₂ - x₁)² + (y₂ - y₁)²] Step 1: Identify coordinates x₁ = 1, y₁ = 3 x₂ = 4, y₂ = 7 Step 2: Calculate differences x₂ - x₁ = 4 - 1 = 3 y₂ - y₁ = 7 - 3 = 4 Step 3: Square the differences (x₂ - x₁)² = 3² = 9 (y₂ - y₁)² = 4² = 16 Step 4: Sum and take square root d = √(9 + 16) = √25 = 5 Answer: Distance PQ = 5 units

Q3: Find the coordinates of the midpoint of the line segment joining A(-2, 4) and B(6, -2).

Points: A(-2, 4) and B(6, -2) Use midpoint formula: M = [(x₁ + x₂)/2, (y₁ + y₂)/2] Step 1: Identify coordinates x₁ = -2, y₁ = 4 x₂ = 6, y₂ = -2 Step 2: Calculate x-coordinate of midpoint x_M = (x₁ + x₂)/2 = (-2 + 6)/2 = 4/2 = 2 Step 3: Calculate y-coordinate of midpoint y_M = (y₁ + y₂)/2 = (4 + (-2))/2 = 2/2 = 1 Step 4: Write midpoint coordinates M = (2, 1) Answer: The midpoint is (2, 1)

Q4: Show that the points A(0, 0), B(3, 4), and C(6, 8) are collinear.

Points: A(0, 0), B(3, 4), C(6, 8) Points are collinear if they lie on the same line. Method 1: Using slope Slope between A and B: m_AB = (4 - 0)/(3 - 0) = 4/3 Slope between B and C: m_BC = (8 - 4)/(6 - 3) = 4/3 Slope between A and C: m_AC = (8 - 0)/(6 - 0) = 8/6 = 4/3 Since all three slopes are equal (4/3), the points are collinear. Method 2: Using area of triangle If area of triangle ABC = 0, then points are collinear. Area = (1/2)|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)| = (1/2)|0(4 - 8) + ...

Q5: Find the area of a triangle with vertices at A(1, 2), B(4, 2), and C(4, 5).

Vertices: A(1, 2), B(4, 2), C(4, 5) Method 1: Using coordinate formula Area = (1/2)|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)| Step 1: Identify coordinates x₁ = 1, y₁ = 2 x₂ = 4, y₂ = 2 x₃ = 4, y₃ = 5 Step 2: Substitute in formula Area = (1/2)|1(2 - 5) + 4(5 - 2) + 4(2 - 2)| = (1/2)|1(-3) + 4(3) + 4(0)| = (1/2)|-3 + 12 + 0| = (1/2)|9| = 9/2 = 4.5 Method 2: Geometric observation Base AB = distance from A(1,2) to B(4,2) = 4 - 1 = 3 Height = perpendicular distance from C(4,5) to line AB = 5 - 2 =...

Q6: A point lies on the x-axis. If its distance from (2, 3) is 5 units, find the coordinates of the point.

Given: Point lies on x-axis, distance from (2, 3) is 5 Step 1: Represent the point Since point is on x-axis, coordinates are (x, 0) for some value of x. Step 2: Apply distance formula Distance = √[(x - 2)² + (0 - 3)²] = 5 Step 3: Square both sides (x - 2)² + 9 = 25 Step 4: Simplify (x - 2)² = 25 - 9 (x - 2)² = 16 Step 5: Take square root x - 2 = ±4 x = 2 + 4 or x = 2 - 4 x = 6 or x = -2 Step 6: Find both points Point 1: (6, 0) Point 2: (-2, 0) Step 7: Verify Distance from (6, 0) to (2, 3)...

Showing 6 of 8 questions. Visit the full page for complete solutions.