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.
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…
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NCERT Solutions for Class 9 Mathematics — Coordinate Geometry. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Coordinate Geometry — All 8 Questions Solved
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, 4 units up
- B is 2 units left, 5 units up
- C is 3 units left, 2 units down
- D is 4 units right, 1 unit down
Answer:
A: Quadrant I, B: Quadrant II, C: Quadrant III, D: Quadrant IV
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) + 3(8 - 0) + 6(0 - 4)|
= (1/2)|0 + 24 - 24|
= (1/2)|0| = 0
Since area = 0, the points are collinear.
Conclusion: A, B, and C are collinear.
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 = 3
Area = (1/2) × base × height = (1/2) × 3 × 3 = 4.5
Answer: Area of triangle = 4.5 square units
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) = √[(6-2)² + (0-3)²] = √[16 + 9] = √25 = 5 ✓
Distance from (-2, 0) to (2, 3) = √[(-2-2)² + (0-3)²] = √[16 + 9] = √25 = 5 ✓
Answer: The points are (6, 0) and (-2, 0)
Q7. Find the slope of the line passing through P(-1, 2) and Q(3, -4). Is the line rising, falling, or horizontal?
Points: P(-1, 2) and Q(3, -4)
Step 1: Apply slope formula
Slope m = (y₂ - y₁)/(x₂ - x₁) = (-4 - 2)/(3 - (-1))
Step 2: Calculate numerator
= (-6)/(3 + 1)
= -6/4
Step 3: Simplify
= -3/2
Step 4: Interpret the slope
Slope m = -3/2 < 0 (negative)
When slope is negative, the line is falling (decreasing from left to right).
Interpretation:
For every 2 units moved to the right, the line drops 3 units down.
Answer: Slope = -3/2; The line is falling.
Q8. Verify that the quadrilateral with vertices A(0, 0), B(4, 0), C(4, 3), D(0, 3) is a rectangle.
Vertices: A(0, 0), B(4, 0), C(4, 3), D(0, 3)
Step 1: Calculate all four sides using distance formula
AB = √[(4-0)² + (0-0)²] = √16 = 4
BC = √[(4-4)² + (3-0)²] = √9 = 3
CD = √[(0-4)² + (3-3)²] = √16 = 4
DA = √[(0-0)² + (0-3)²] = √9 = 3
Step 2: Check opposite sides are equal
AB = CD = 4 ✓
BC = DA = 3 ✓
Opposite sides are equal.
Step 3: Check all angles are right angles using slopes
Slope of AB = (0-0)/(4-0) = 0 (horizontal)
Slope of BC = (3-0)/(4-4) = undefined (vertical)
Slope of CD = (3-3)/(0-4) = 0 (horizontal)
Slope of DA = (0-3)/(0-0) = undefined (vertical)
Step 4: Verify perpendicularity
AB is perpendicular to BC (horizontal ⊥ vertical) ✓
BC is perpendicular to CD (vertical ⊥ horizontal) ✓
All angles are 90°.
Step 5: Calculate diagonals
AC = √[(4-0)² + (3-0)²] = √[16 + 9] = √25 = 5
BD = √[(0-4)² + (3-0)²] = √[16 + 9] = √25 = 5
Diagonals are equal: AC = BD ✓
Conclusion: ABCD is a rectangle (opposite sides equal, all angles 90°, diagonals equal).
Coordinate Geometry — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Point P(-3, 5) lies in which quadrant?
- A. I
- B. II
- C. III
- D. IV
Answer: B. II — In the second quadrant, the x-coordinate is negative and the y-coordinate is positive. For P(-3, 5), x = -3 (negative) and y = 5 (positive).
MCQ 2. The abscissa of the point (-7, 2) is:
- A. -7
- B. 2
- C. 0
- D. -5
Answer: A. -7 — The abscissa refers to the x-coordinate of a point. For the point (-7, 2), the x-coordinate is -7.
MCQ 3. The coordinates of the origin are:
- A. (1, 0)
- B. (0, 1)
- C. (0, 0)
- D. (1, 1)
Answer: C. (0, 0) — The origin is the point where the x-axis and y-axis intersect, and its coordinates are always (0, 0).
MCQ 4. A point whose y-coordinate is zero will always lie on which axis?
- A. x-axis
- B. y-axis
- C. origin
- D. either x-axis or y-axis
Answer: A. x-axis — Any point on the x-axis has its y-coordinate as zero, i.e., it is of the form (x, 0).
MCQ 5. A point whose x-coordinate is zero will always lie on which axis?
- A. x-axis
- B. y-axis
- C. origin
- D. either x-axis or y-axis
Answer: B. y-axis — Any point on the y-axis has its x-coordinate as zero, i.e., it is of the form (0, y).
MCQ 6. The perpendicular distance of the point (4, -3) from the x-axis is:
- A. 4
- B. -3
- C. 3
- D. 7
Answer: C. 3 — The perpendicular distance of a point (x, y) from the x-axis is given by the absolute value of its y-coordinate, which is |-3| = 3 units.
MCQ 7. The perpendicular distance of the point (-5, 6) from the y-axis is:
- A. -5
- B. 5
- C. 6
- D. 11
Answer: B. 5 — The perpendicular distance of a point (x, y) from the y-axis is given by the absolute value of its x-coordinate, which is |-5| = 5 units.
MCQ 8. The signs of the coordinates of a point in the third quadrant are:
- A. (+, +)
- B. (-, +)
- C. (-, -)
- D. (+, -)
Answer: C. (-, -) — In the third quadrant, both the x-coordinate and the y-coordinate are negative.
MCQ 9. If the points A(2,0), B(0,2), C(-2,0) and D(0,-2) are plotted on a graph paper, what figure do they form?
- A. Square
- B. Rectangle
- C. Rhombus
- D. Parallelogram
Answer: A. Square — Plotting these points reveals that all four sides are equal in length and diagonals are equal and bisect each other perpendicularly, forming a square.
MCQ 10. Which of the following points lie on a line parallel to the x-axis?
- A. (3, 2), (3, -1), (3, 0)
- B. (1, 4), (2, 4), (-5, 4)
- C. (1, 1), (2, 2), (3, 3)
- D. (0, 0), (1, 0), (0, 1)
Answer: B. (1, 4), (2, 4), (-5, 4) — A line parallel to the x-axis has a constant y-coordinate. In option B, all points have y-coordinate 4.
MCQ 11. A point for which its abscissa and ordinate are equal lies on the line:
- A. x = 0
- B. y = 0
- C. y = x
- D. y = -x
Answer: C. y = x — If the abscissa (x-coordinate) and ordinate (y-coordinate) are equal, then x = y, which is the equation of the line passing through the origin with a slope of 1.
MCQ 12. A point for which its ordinate is the negative of its abscissa lies on the line:
- A. x = 0
- B. y = 0
- C. y = x
- D. y = -x
Answer: D. y = -x — If the ordinate (y-coordinate) is the negative of its abscissa (x-coordinate), then y = -x.
MCQ 13. A point is 5 units away from the y-axis and 3 units away from the x-axis. If it lies in the fourth quadrant, its coordinates are:
- A. (5, 3)
- B. (-5, 3)
- C. (-5, -3)
- D. (5, -3)
Answer: D. (5, -3) — The distance from the y-axis is the x-coordinate, and from the x-axis is the y-coordinate. In the fourth quadrant, x is positive and y is negative, so the point is (5, -3).
MCQ 14. The point (0, -7) lies:
- A. on the x-axis
- B. in the first quadrant
- C. on the y-axis
- D. in the fourth quadrant
Answer: C. on the y-axis — Since the x-coordinate is 0, the point lies on the y-axis.
MCQ 15. If the coordinates of two points are P(2, -3) and Q(-3, 2), then:
- A. The abscissa of P is equal to the ordinate of Q.
- B. The ordinate of P is equal to the abscissa of Q.
- C. The abscissa of P is equal to the abscissa of Q.
- D. The ordinate of P is equal to the ordinate of Q.
Answer: B. The ordinate of P is equal to the abscissa of Q. — The ordinate of P is -3 and the abscissa of Q is -3. Hence, they are equal.
MCQ 16. The points A(3, 4), B(3, 0), C(3, -2) lie on a line:
- A. parallel to x-axis
- B. parallel to y-axis
- C. passing through the origin
- D. none of these
Answer: B. parallel to y-axis — All the points have the same x-coordinate (which is 3). This means they lie on a vertical line x=3, which is parallel to the y-axis.
MCQ 17. What is the area of the triangle formed by the points O(0,0), A(5,0) and B(0,3)?
- A. 15 sq units
- B. 7.5 sq units
- C. 8 sq units
- D. 10 sq units
Answer: B. 7.5 sq units — This is a right-angled triangle with base OA along the x-axis (length 5 units) and height OB along the y-axis (length 3 units). Area = 1/2 × base × height = 1/2 × 5 × 3 = 7.5 sq units.
MCQ 18. If the point (k, 3) lies on the line passing through (2, 3) and (5, 3), then the value of k is:
- A. 2
- B. 5
- C. 3
- D. any real number
Answer: D. any real number — All points (2, 3) and (5, 3) have the same y-coordinate, 3. This means they lie on the horizontal line y=3. If (k, 3) also lies on this line, then k can be any real number.
MCQ 19. If the vertices of a square are (0,0), (a,0), (a,a), and (0,a) where a > 0, then its perimeter is:
- A. a
- B. 2a
- C. 4a
- D. a^2
Answer: C. 4a — The side length of the square is 'a' units (e.g., the distance between (0,0) and (a,0) is 'a'). A square has four equal sides, so its perimeter is 4 × side = 4a.
MCQ 20. A point (x, y) such that x > 0, y < 0 and |x| > |y| lies in which quadrant?
- A. I
- B. II
- C. III
- D. IV
Answer: D. IV — If x > 0 and y < 0, the point lies in the fourth quadrant. The condition |x| > |y| further specifies its position but does not change the quadrant itself.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.