Class 10 Maths Chapter 7 Coordinate Geometry Practice Paper 2

PRACTICE PAPER 02 - CHAPTER 07 COORDINATE GEOMETRY (2025-26) SUBJECT: MATHEMATICS MAX. MARKS: 40 CLASS: X DURATION: 1½ hrs

General Instructions:

1. All questions are compulsory.

2. This question paper contains 20 questions divided into five Sections A, B, C, D and E.

3. Section A comprises of 10 MCQs of 1 mark each. Section B comprises of 4 questions of 2 marks each.

Section C comprises of 3 questions of 3 marks each. Section D comprises of 1 question of 5 marks

and Section E comprises of 2 Case Study Based Questions of 4 marks each.

4. There is no overall choice.

5. Use of Calculators is not permitted. SECTION – A Questions 1 to 10 carry 1 mark each.

1. The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is:
(a) 5
(b) 12
(c) 11
(d) 7 + √5

2. If the coordinates of the mid-points of the sides of a triangle are (1, 1), (2, –3) and (3, 4), then the centroid is:
(a) (3, 1)
(b) (2, 2/3)
(c) (2, 1)
(d) (2, 2)

3. If the points (k, 2k), (3k, 3k) and (3, 1) are collinear, then k is:
(a) 1/3
(b) –1/3
(c) 2/3
(d) –2/3

4. The area of a rhombus whose vertices are (3, 0), (4, 5), (–1, 4) and (–2, –1) is:
(a) 12 sq. units
(b) 24 sq. units
(c) 30 sq. units
(d) 32 sq. units

5. The fourth vertex D of parallelogram ABCD with A(–2, 3), B(6, 7) and C(8, 3) is:
(a) (0, 1)
(b) (0, –1)
(c) (–1, 0)
(d) (1, 0)

6. If P(a/3, 4) is the mid-point of line segment joining Q(–6, 5) and R(–2, 3), then a is:
(a) –4
(b) –12
(c) 12
(d) –6

7. If P(x, y) is equidistant from A(a+b, b–a) and B(a–b, a+b), then:
(a) ax = by
(b) bx = ay
(c) ax + by = 0
(d) bx – ay = 0

8. The point dividing (7, –6) and (3, 4) in ratio 1:2 internally lies in:
(a) I quadrant
(b) II quadrant
(c) III quadrant
(d) IV quadrant In questions 9 and 10, choose the correct answer:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true

9. Assertion
(a) : Points (1, 2), (–1, –16) and (0, –7) lie on a straight line. Reason (R): Three points are collinear if area of triangle is zero.

10. Assertion
(a) : Point P(0, 2) is intersection of y-axis with line 3x + 2y = 4. Reason (R): Distance of P(0, 2) from x-axis is 2 units. SECTION – B Questions 11 to 14 carry 2 marks each.

11. Find values of k for which points (8, 1), (3, –2k) and (k, –5) are collinear.

12. Show that quadrilateral formed by mid-points of consecutive sides of a square is also a square.

13. Find coordinates of points dividing line segment joining A(–2, 2) and B(2, 8) into four equal parts.

14. If A(–5, 7), B(–4, –5), C(–1, –6) and D(4, 5) are vertices of a quadrilateral, find its area. SECTION – C Questions 15 to 17 carry 3 marks each.

15. Find coordinates of points of trisection of line segment joining A(2, –2) and B(–7, 4).

16. If vertices of triangle are (1, –3), (4, p) and (–9, 7) and area is 15 sq. units, find p.

17. Prove that points (2, –1), (0, 2), (3, 3) and (5, 0) are vertices of a parallelogram. OR If (1, 2), (4, y), (x, 6) and (3, 5) are vertices of a parallelogram, find x and y. SECTION – D Question 18 carries 5 marks.

18. Find area of quadrilateral with vertices (–4, –2), (–3, –5), (3, –2) and (2, 3). OR Prove that (4, 3), (6, 4), (5, 6) and (3, 5) are vertices of a square. Find its area. SECTION – E (Case Study Based Questions) Questions 19 to 20 carry 4 marks each.

19. TREASURE HUNT Three treasures at A(–2, 4), B(2, –3), C(5, 2). Starting point S(0, 0). (i) Which treasure is closest to S? Find distance. (1 mark) (ii) Find coordinates equidistant from all three treasures. (2 marks) OR Find total distance from A to B to C. (2 marks) (iii) Bonus treasure T divides AC in ratio 2:3. Find coordinates of T. (1 mark)

20. MOBILE TOWERS Villages at P(4, 6), Q(8, 10), R(12, 2). Units in km. (i) Find distance PQ. (1 mark) (ii)
(a) Find midpoint M of PQ. (1 mark) OR (ii)
(b) Find area of triangle PQR. (1 mark) (iii) Find centroid of triangle PQR. (2 marks) DETAILED ANSWER KEY

SECTION A – ANSWERS

1.
(b) 12 AB=4, BC=3, AC=5. Perimeter = 4+3+5 = 12

2.
(d) (2, 2) Centroid of midpoints = ((1+2+3)/3, (1–3+4)/3) = (2, 2/3). Using property of midpoints: actual centroid is (2, 2)

3.
(b) –1/3 For collinearity, area = 0. Solving gives k = –1/3

4.
(b) 24 sq. units Area = (1/2) × d₁ × d₂ = (1/2) × 4√2 × 6√2 = 24

5.
(b) (0, –1) Diagonals bisect. Midpoint of AC = Midpoint of BD. Solving: D(0, –1)

6.
(b) –12 a/3 = (–6–2)/2 = –4, so a = –12

7.
(b) bx = ay Equal distances gives bx = ay

8.
(d) IV quadrant Point = (6, –5). x>0, y<0 → IV quadrant

9.
(a) Both true, R explains A

10.
(a) Both true, R explains A

SECTION B – ANSWERS

11. k = 2 or k = 11/2 Using collinearity condition: 2k²–15k+22=0. Solutions: k=2, k=11/2

12. All sides equal, diagonals equal → Square Take square with vertices at (0,0), (a,0), (a,a), (0,a). Find midpoints. Prove all sides = a/√2

13. (–1, 7/2), (0, 5), (1, 11/2) Divide in ratios 1:3, 1:1, 3:1

14. 72 sq. units Area = Area(ABC) + Area(ACD) = 35/2 + 109/2 = 72

SECTION C – ANSWERS

15. (–1, 0) and (–4, 2) Points divide in ratios 1:2 and 2:1

16. p = –3 or p = –9 Area = 15. Using formula: |10p+60| = 30. So p = –3 or –9

17. Opposite sides equal → Parallelogram AB=√13=CD, BC=√10=AD OR: x = 6, y = 3 Diagonals bisect each other. Solving gives x=6, y=3

SECTION D – ANSWER

18. 28 sq. units Area = Area(ABC) + Area(ACD) = 21/2 + 35/2 = 28 OR: All sides = √5, diagonals equal, perpendicular. Area = 5 sq. units AB=BC=CD=DA=√5. AC=BD=√10. Area = (√5)² = 5

SECTION E – ANSWERS

19. (i) B closest: √13 ≈ 3.61 units (ii) Circumcenter ≈ (0.94, 1.04) OR: Total = √65 + √34 ≈ 13.89 units (iii) T = (4/5, 16/5) 20. (i) PQ = 4√2 km (ii)
(a) M = (6, 8) OR (ii)
(b) Area = 24 sq. km (iii) Centroid = (8, 6)