Coordinate Geometry — Class 10 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Coordinate Geometry" — 9 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Coordinate Geometry" — 9 important questions with detailed answers for CBSE board…
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NCERT Solutions for Class 10 Mathematics — Coordinate Geometry. Step-by-step answers to all 9 textbook questions from this chapter, written for CBSE board preparation and free to use.
Coordinate Geometry — All 9 Questions Solved
Q1. State the distance formula. Find the distance between points A(3, 4) and B(6, 8).
Distance Formula:
The distance d between two points P(x₁, y₁) and Q(x₂, y₂) in a coordinate plane is:
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
This formula is derived from the Pythagorean theorem.
Problem: Find distance between A(3, 4) and B(6, 8)
Step 1: Identify the coordinates.
Point A: x₁ = 3, y₁ = 4
Point B: x₂ = 6, y₂ = 8
Step 2: Substitute into the distance formula.
d = √[(6 - 3)² + (8 - 4)²]
Step 3: Simplify inside the square root.
d = √[(3)² + (4)²]
d = √[9 + 16]
d = √25
d = 5 units
Answer: Distance AB = 5 units
Q2. Find the coordinates of the point that divides the line segment joining A(2, 3) and B(8, 9) in the ratio 2:1 internally.
Section Formula (Internal Division):
If point P divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m:n, then:
P = ((m·x₂ + n·x₁)/(m + n), (m·y₂ + n·y₁)/(m + n))
Given:
A(2, 3), B(8, 9)
Ratio = 2:1 (m:n = 2:1)
Step 1: Identify the values.
x₁ = 2, y₁ = 3
x₂ = 8, y₂ = 9
m = 2, n = 1
Step 2: Apply the section formula for x-coordinate.
x = (m·x₂ + n·x₁)/(m + n)
x = (2·8 + 1·2)/(2 + 1)
x = (16 + 2)/3
x = 18/3
x = 6
Step 3: Apply the section formula for y-coordinate.
y = (m·y₂ + n·y₁)/(m + n)
y = (2·9 + 1·3)/(2 + 1)
y = (18 + 3)/3
y = 21/3
y = 7
Answer: The point is P(6, 7)
Verification: Point (6, 7) lies on AB and divides it in ratio 2:1.
Q3. Prove that the points A(0, 0), B(3, 4), and C(6, 8) are collinear.
Given: Points A(0, 0), B(3, 4), C(6, 8)
Prove: These points are collinear (lie on the same line)
Method 1: Using Slope
Step 1: Find the slope of AB.
Slope of AB = (y₂ - y₁)/(x₂ - x₁) = (4 - 0)/(3 - 0) = 4/3
Step 2: Find the slope of BC.
Slope of BC = (8 - 4)/(6 - 3) = 4/3
Step 3: Check if slopes are equal.
Slope of AB = Slope of BC = 4/3
Also, B lies on both lines, and A and C are on the same side of B.
Step 4: Conclusion.
Since the slopes are equal and B is a common point, the three points are collinear.
Method 2: Using Distance Formula
Step 1: Find AB.
AB = √[(3 - 0)² + (4 - 0)²] = √[9 + 16] = √25 = 5
Step 2: Find BC.
BC = √[(6 - 3)² + (8 - 4)²] = √[9 + 16] = √25 = 5
Step 3: Find AC.
AC = √[(6 - 0)² + (8 - 0)²] = √[36 + 64] = √100 = 10
Step 4: Check the condition.
AB + BC = 5 + 5 = 10 = AC ✓
Since AB + BC = AC, point B lies on the line segment AC.
Therefore, A, B, and C are collinear. ✓
Q4. Find the area of the triangle with vertices A(1, 2), B(4, 6), and C(5, 1).
Area of Triangle using Coordinate Formula:
For a triangle with vertices A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃):
Area = (1/2)|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
Given:
A(1, 2), B(4, 6), C(5, 1)
x₁ = 1, y₁ = 2
x₂ = 4, y₂ = 6
x₃ = 5, y₃ = 1
Step 1: Substitute into the formula.
Area = (1/2)|1(6 - 1) + 4(1 - 2) + 5(2 - 6)|
Step 2: Calculate each term.
= (1/2)|1(5) + 4(-1) + 5(-4)|
= (1/2)|5 - 4 - 20|
= (1/2)|-19|
= (1/2)(19)
= 9.5 square units
Answer: Area of triangle ABC = 9.5 square units or 19/2 square units
Alternative method verification using the formula:
Area = (1/2)|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
We can also write this as:
Area = (1/2)|x₁y₂ - x₁y₃ + x₂y₃ - x₂y₁ + x₃y₁ - x₃y₂|
= (1/2)|1(6) - 1(1) + 4(1) - 4(2) + 5(2) - 5(6)|
= (1/2)|6 - 1 + 4 - 8 + 10 - 30|
= (1/2)|-19| = 9.5 ✓
Q5. The midpoint of a line segment joining A and B is M(2, 3). If A is (0, 1), find the coordinates of B.
Midpoint Formula:
If M is the midpoint of AB where A(x₁, y₁) and B(x₂, y₂), then:
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Given:
M(2, 3) is the midpoint
A(0, 1)
Find: B(x₂, y₂)
Step 1: Apply the midpoint formula for the x-coordinate.
2 = (0 + x₂)/2
4 = 0 + x₂
x₂ = 4
Step 2: Apply the midpoint formula for the y-coordinate.
3 = (1 + y₂)/2
6 = 1 + y₂
y₂ = 5
Answer: B(4, 5)
Verification:
Midpoint of A(0, 1) and B(4, 5) = ((0 + 4)/2, (1 + 5)/2) = (2, 3) ✓
Q6. Three vertices of a parallelogram are A(1, 2), B(4, 3), and C(6, 6). Find the coordinates of the fourth vertex D.
Property of Parallelogram:
In a parallelogram, the diagonals bisect each other. If ABCD is a parallelogram with diagonals AC and BD, then the midpoint of AC equals the midpoint of BD.
Given:
A(1, 2), B(4, 3), C(6, 6)
Find: D(x, y)
Assuming the vertices are in order A, B, C, D:
Step 1: Find the midpoint of diagonal AC.
Midpoint of AC = ((1 + 6)/2, (2 + 6)/2) = (7/2, 4) = (3.5, 4)
Step 2: This midpoint must equal the midpoint of diagonal BD.
Midpoint of BD = ((4 + x)/2, (3 + y)/2)
Step 3: Set the midpoints equal.
(4 + x)/2 = 7/2
(3 + y)/2 = 4
Step 4: Solve for x.
4 + x = 7
x = 3
Step 5: Solve for y.
3 + y = 8
y = 5
Answer: D(3, 5)
Verification:
Midpoint of AC = (3.5, 4)
Midpoint of BD = ((4 + 3)/2, (3 + 5)/2) = (7/2, 8/2) = (3.5, 4) ✓
The diagonals bisect each other, confirming ABCD is a parallelogram.
Q7. Find the distance of the point A(5, -3) from the origin O(0, 0).
Given: Point A(5, -3), Origin O(0, 0)
Find: Distance OA
Step 1: Apply the distance formula.
Distance = √[(x₂ - x₁)² + (y₂ - y₁)²]
Step 2: Substitute values.
OA = √[(5 - 0)² + (-3 - 0)²]
OA = √[(5)² + (-3)²]
OA = √[25 + 9]
OA = √34 units
Answer: Distance from A to origin = √34 units ≈ 5.83 units
Q8. The points A(-1, 2), B(3, -4), and C(5, -2) form a triangle. Find the length of the median from A to the midpoint of BC.
Median:
A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side.
Given:
A(-1, 2), B(3, -4), C(5, -2)
Find: Length of median from A to midpoint of BC
Step 1: Find the midpoint of BC.
Let M be the midpoint of BC.
M = ((3 + 5)/2, (-4 + (-2))/2)
M = (8/2, -6/2)
M = (4, -3)
Step 2: Find the distance from A to M (this is the median).
AM = √[(4 - (-1))² + (-3 - 2)²]
AM = √[(4 + 1)² + (-5)²]
AM = √[(5)² + (-5)²]
AM = √[25 + 25]
AM = √50
AM = 5√2 units
Answer: Length of median from A = 5√2 units ≈ 7.07 units
Q9. Show that the points P(6, -1), Q(5, 0), and R(4, 1) are collinear.
Given: Points P(6, -1), Q(5, 0), R(4, 1)
Show: The points are collinear
Method: Using the Area Formula
Three points are collinear if the area of the triangle formed by them is zero.
Step 1: Apply the area formula.
Area = (1/2)|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
Step 2: Substitute the coordinates.
With P(6, -1), Q(5, 0), R(4, 1):
Area = (1/2)|6(0 - 1) + 5(1 - (-1)) + 4((-1) - 0)|
Area = (1/2)|6(-1) + 5(2) + 4(-1)|
Area = (1/2)|-6 + 10 - 4|
Area = (1/2)|0|
Area = 0
Step 3: Conclusion.
Since the area is zero, the points P, Q, and R are collinear.
Alternative verification using slopes:
Slope of PQ = (0 - (-1))/(5 - 6) = 1/(-1) = -1
Slope of QR = (1 - 0)/(4 - 5) = 1/(-1) = -1
Slope of PR = (1 - (-1))/(4 - 6) = 2/(-2) = -1
Since all slopes are equal and the points are connected, P, Q, and R are collinear. ✓
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. The distance of the point P(3, 4) from the origin is:
- A. 3
- B. 4
- C. 5
- D. 7
Answer: C. 5 — Use the distance formula from the origin (0,0), which is √(x² + y²). So, √(3² + 4²) = √(9 + 16) = √25 = 5.
MCQ 2. The midpoint of the line segment joining the points A(2, 3) and B(4, 7) is:
- A. (3, 5)
- B. (6, 10)
- C. (1, 2)
- D. (2, 3)
Answer: A. (3, 5) — The midpoint formula is ((x₁ + x₂)/2, (y₁ + y₂)/2). Applying this, we get ((2+4)/2, (3+7)/2) = (6/2, 10/2) = (3, 5).
MCQ 3. A point on the x-axis has coordinates of the form:
- A. (0, y)
- B. (x, 0)
- C. (x, y)
- D. (0, 0)
Answer: B. (x, 0) — Any point located on the x-axis will always have its y-coordinate equal to 0, while its x-coordinate can be any real number.
MCQ 4. The distance between the points (a cos θ, 0) and (0, a sin θ) is:
- A. a
- B. a²
- C. 2a
- D. 1
Answer: A. a — Apply the distance formula √((x₂-x₁)² + (y₂-y₁)²). This simplifies to √(a²cos²θ + a²sin²θ) = √(a²(cos²θ + sin²θ)) = √(a² * 1) = a.
MCQ 5. The ratio in which the y-axis divides the line segment joining the points A(5, -6) and B(-1, -4) is:
- A. 1:5
- B. 5:1
- C. 1:1
- D. 2:3
Answer: B. 5:1 — A point on the y-axis has its x-coordinate equal to 0. Using the section formula (mx₂+nx₁)/(m+n) for the x-coordinate, (k*(-1) + 1*5)/(k+1) = 0, which gives -k+5=0, so k=5. The ratio is 5:1.
MCQ 6. The area of the triangle formed by the points (0,0), (a,0), and (0,b) is:
- A. ab
- B. ½ab
- C. 2ab
- D. √(a²+b²)
Answer: B. ½ab — This triangle is a right-angled triangle with base 'a' along the x-axis and height 'b' along the y-axis. Its area is given by ½ × base × height, which is ½ab.
MCQ 7. If the points A(x, 2), B(-3, -4) and C(7, -5) are collinear, then the value of x is:
- A. -63
- B. 63
- C. -3
- D. 3
Answer: A. -63 — For three points to be collinear, the slope between any two pairs of points must be equal. Equating slope AB and slope BC gives (-4-2)/(-3-x) = (-5-(-4))/(7-(-3)), leading to x = -63.
MCQ 8. The vertices of a triangle are (0,0), (3,0), and (0,4). What type of triangle is it?
- A. Equilateral
- B. Isosceles
- C. Right-angled
- D. Scalene
Answer: C. Right-angled — The sides have lengths 3 (between (0,0) and (3,0)), 4 (between (0,0) and (0,4)), and 5 (between (3,0) and (0,4), using distance formula). Since 3² + 4² = 5² (9+16=25), it satisfies the Pythagorean theorem, making it a right-angled triangle.
MCQ 9. Find the coordinates of the point which divides the join of P(-1, 7) and Q(4, -3) in the ratio 2:3 internally.
- A. (1, 3)
- B. (2, 4)
- C. (3, 5)
- D. (0, 1)
Answer: A. (1, 3) — Use the section formula (mx₂ + nx₁)/(m+n) and (my₂ + ny₁)/(m+n). For x: (2*4 + 3*(-1))/(2+3) = 5/5 = 1. For y: (2*(-3) + 3*7)/(2+3) = 15/5 = 3. So the point is (1, 3).
MCQ 10. The distance of the point (2, 3) from the x-axis is:
- A. 2
- B. 3
- C. √13
- D. 1
Answer: B. 3 — The distance of a point (x, y) from the x-axis is the absolute value of its y-coordinate, which is |3| = 3 units.
MCQ 11. The distance of the point (2, 3) from the y-axis is:
- A. 2
- B. 3
- C. √13
- D. 1
Answer: A. 2 — The distance of a point (x, y) from the y-axis is the absolute value of its x-coordinate, which is |2| = 2 units.
MCQ 12. If the midpoint of the line segment joining the points A(3, 4) and B(k, 6) is P(x, y) and x+y-10=0, then the value of k is:
- A. 1
- B. 3
- C. 5
- D. 7
Answer: D. 7 — The midpoint P is ((3+k)/2, (4+6)/2) = ((3+k)/2, 5). Substitute these coordinates into x+y-10=0: ((3+k)/2) + 5 - 10 = 0. This simplifies to (3+k)/2 = 5, so 3+k = 10, which means k=7.
MCQ 13. The area of a triangle with vertices (1, -1), (-4, 6) and (-3, -5) is:
- A. 24 sq units
- B. 48 sq units
- C. 12 sq units
- D. 36 sq units
Answer: A. 24 sq units — Using the area formula ½ |x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|, we get ½ |1(6 - (-5)) + (-4)(-5 - (-1)) + (-3)(-1 - 6)| = ½ |11 + 16 + 21| = ½ |48| = 24 sq units.
MCQ 14. The coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) such that AP = 2/5 AB are:
- A. (3, 4)
- B. (2, 3)
- C. (4, 5)
- D. (3.5, 4.5)
Answer: A. (3, 4) — If AP = 2/5 AB, then the point P divides the line segment AB in the ratio 2:3. Using the section formula, x = (2*6 + 3*1)/(2+3) = 15/5 = 3 and y = (2*7 + 3*2)/(2+3) = 20/5 = 4. Thus, P is (3, 4).
MCQ 15. The points (2, 1), (0, 0), (-1, 2) and (1, 3) taken in order are the vertices of a:
- A. Rectangle
- B. Rhombus
- C. Parallelogram
- D. Square
Answer: D. Square — Let the points be A(2,1), B(0,0), C(-1,2), D(1,3). Calculating distances: AB=BC=CD=DA=√5 (all sides equal). Calculating diagonals: AC=BD=√10 (diagonals equal). Since all sides are equal and diagonals are equal, it's a Square.
MCQ 16. If the centroid of a triangle with vertices (3, -5), (-7, 4), and (10, -2) is (x, y), find x+y.
- A. 0
- B. 1
- C. 2
- D. -1
Answer: B. 1 — The centroid G is ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3). So, G = ((3-7+10)/3, (-5+4-2)/3) = (6/3, -3/3) = (2, -1). Therefore, x=2, y=-1, and x+y = 2+(-1) = 1.
MCQ 17. The perpendicular distance of the point (4, 3) from the x-axis is:
- A. 4
- B. 3
- C. 5
- D. 1
Answer: B. 3 — The perpendicular distance of a point (x, y) from the x-axis is the absolute value of its y-coordinate, which is |3| = 3 units.
MCQ 18. Which of the following points lies on the line joining (1, -1) and (4, 5)?
- A. (2, 1)
- B. (3, 2)
- C. (5, 4)
- D. (6, 7)
Answer: A. (2, 1) — The slope of the line is (5 - (-1))/(4 - 1) = 6/3 = 2. The equation of the line is y - (-1) = 2(x - 1), which simplifies to y = 2x - 3. Checking the options, only (2, 1) satisfies this equation (1 = 2*2 - 3).
MCQ 19. The vertices of a triangle are A(2, 3), B(-1, 0), and C(k, -4). If the area of triangle ABC is 12 sq units, then find the value of k.
- A. 3
- B. 5
- C. -3
- D. -5
Answer: A. 3 — Using the area formula ½ |x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)| = 12. Substituting values: ½ |2(0-(-4)) + (-1)(-4-3) + k(3-0)| = 12. This simplifies to ½ |8 + 7 + 3k| = 12, or |15 + 3k| = 24. This gives 15 + 3k = 24 (k=3) or 15 + 3k = -24 (k=-13). Among the options, 3 is correct.
MCQ 20. The ratio in which the point (11, y) divides the line segment joining P(15, 5) and Q(9, 20) is:
- A. 2:1
- B. 1:2
- C. 3:1
- D. 1:3
Answer: A. 2:1 — Let the ratio be k:1. Using the section formula for the x-coordinate: (k*9 + 1*15)/(k+1) = 11. This simplifies to 9k + 15 = 11k + 11, which gives 2k = 4, so k=2. The ratio is 2:1.
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.