Class 10 Maths Chapter 7 Coordinate Geometry Practice Paper 1

Q1. All questions are compulsory.

Answer: (a) ±4
Distance = √[(4–1)² + p²] = 5 → 9+p² = 25 → p = ±4

Q2. This question paper contains 20 questions divided into five Sections A, B, C, D and E.

Answer: (d) (16/3, 6)
Centroid = ((0+8+8)/3, (6+12+0)/3) = (16/3, 6)

Q3. 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.

Answer: (a) 1
For ∠B=90°: AB²+BC²=AC². After calculation, t = 1

Q4. There is no overall choice.

Answer: (a) 1 : 4
Using section formula and solving, ratio = 1:4

Q5. Use of Calculators is not permitted. SECTION – A Questions 1 to 10 carry 1 mark each.
1. If the distance between points (4, p) and (1, 0) is 5 units, then the value of p is:
(a) ±4 (b) ±3 (c) 4 only (d) 0
2. The coordinates of the centroid of a triangle whose vertices are (0, 6), (8, 12) and (8, 0) is:
(a) (4, 6) (b) (16, 6) (c) (8, 6) (d) (16/3, 6)
3. If A(5, 2), B(2, –2) and C(–2, t) are the vertices of a right angled triangle with ∠B = 90°, then the value of t is:
(a) 1 (b) 2 (c) 3 (d) 4
4. The ratio in which the point P(3/4, 5/12) divides the line segment joining the points A(1/2, 3/2) and B(2, –5) is:
(a) 1 : 4 (b) 1 : 5 (c) 2 : 3 (d) 3 : 4
5. If the points (2, 1) and (1, –2) are equidistant from the point (x, y), then:
(a) x + 3y = 0 (b) 3x + y = 0 (c) x – 3y = 0 (d) 3x – y = 0

Answer: (a) x + 3y = 0
Equidistant means equal distances. Solving gives x+3y=0

Q6. The area of triangle formed by the points (a, b+c), (b, c+a) and (c, a+b) is:
(a) 0 (b) a+b+c (c) ab+bc+ca (d) abc

Answer: (a) 0
Area calculation shows points are collinear, so area = 0

Q7. If the point C(–1, 2) divides internally the line segment joining A(2, 5) and B in ratio 3 : 4, then the coordinates of B are:
(a) (–5, –2) (b) (5, 2) (c) (–5, 2) (d) (5, –2)

Answer: (a) (–5, –2)
Using section formula: B = (–5, –2)

Q8. If three points (0, 0), (3, √3) and (3, λ) form an equilateral triangle, then λ is equal to:
(a) 2 (b) –3 (c) –√3 (d) –2√3
In questions 9 and 10, a statement of assertion (A) is followed by a statement of reason (R). 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

Answer: (c) –√3
For equilateral triangle, all sides equal. Solving gives λ = –√3

Q9. Assertion (A): Mid-point of a line segment divides line in the ratio 1 : 1.
Reason (R): The ratio in which the point (–3, k) divides the line segment joining (–5, 4) and (–2, 3) is 1 : 2.

Answer: (a) Both true and R explains A

Q10. Assertion (A): The origin is the only point equidistant from (2, 3) and (-2, -3).
Reason (R): The origin is the mid-point of the line joining (2, 3) and (-2, -3). SECTION – B Questions 11 to 14 carry 2 marks each.

Answer: (a) Both true and R explains A

Q11. The line segment AB joining A(3, –4) and B(1, 2) is trisected at points P(p, –2) and Q(5/3, q). Find the values of p and q.

Answer: p = 7/3, q = 0
Trisection divides in 1:2 and 2:1. Calculate accordingly.

Q12. Find the point on x-axis which is equidistant from the points (2, –5) and (–2, 9).

Answer: (–7, 0)
Let point be (x,0). Equal distances give x = –7

Q13. Find the value of x such that PQ = QR where the coordinates of P, Q and R are (6, –1), (1, 3) and (x, 8)
respectively.

Answer: x = –4
PQ = QR. Distance formula gives x = –4

Q14. Find the coordinates of the points of trisection of the line segment joining (1, –2) and (–3, 4). SECTION – C Questions 15 to 17 carry 3 marks each.

Answer: (1/3, 0) and (–5/3, 2) Trisection points at ratios 1:2 and 2:1

Q15. Show that the points A(3, 5), B(6, 0), C(1, –3) and D(–2, 2) are the vertices of a square ABCD.

Answer: All sides equal, diagonals equal → Square
AB=BC=CD=DA=√34, AC=BD=√68

Q16. In what ratio does the line x – y – 2 = 0 divide the line segment joining (3, –1) and (8, 9)?

Answer: 2:3 Solve using section formula with line equation

Q17. Show that points A(7, 5), B(2, 3) and C(6, –7) are the vertices of a right triangle. Also find its area. OR Find the ratio in which the point (2, y) divides the line segment joining the points A(–2, 2) and B(3, 7). Also find the value of y. SECTION – D Question 18 carries 5 marks.

Answer: Right triangle at B, Area = 32 sq units
AB²+BC²=AC². Area = (1/2)×AB×BC = 32
OR: Ratio 4:1, y = 6
Using section formula: ratio 4:1, y = 6

Q18. Find the centre of a circle passing through (5, –8), (2, –9) and (2, 1). OR If the points (10, 5), (8, 4) and (6, 6) are the mid-points of the sides of a triangle, find its vertices. SECTION – E (Case Study Based Questions) Questions 19 to 20 carry 4 marks each.

Answer: Centre: (3, –4) Equidistant from all three points. Solve system of equations. OR: Vertices: (12, 3), (6, 7), (8, 5) Use midpoint formulas to find original vertices

Q19. GPS NAVIGATION In a GPS system, three locations are marked: Lucknow L(4, 10), Bhuj B(2, 6), and Puri P(10, 6). Distances represent 150 km per unit.
(i) Find the distance between Lucknow and Bhuj. (1 mark)
(ii) If Kota K divides line LB in ratio 3:2, find coordinates of K. (1 mark)
(iii) Find the type of triangle formed by L, B and P. (2 marks) OR Find a point on y-axis equidistant from L and P. (2 marks)

Answer: (i) LB = √[(4–2)²+(10–6)²] = √20 = 2√5 units = 300√5 km
(ii) K = ((3×2+2×4)/5, (3×6+2×10)/5) = (14/5, 38/5)
(iii) Isosceles triangle (two equal sides) OR: Point (0, 8) on y-axis

Q20. FIELD PLANNING A field is shaped as right triangle AQC with square PQRS inside. P(–200, 0), Q(200, 0), and A(200, 800).
(i) What are coordinates of R and S? (1 mark) (ii)(a) What is area of square PQRS? (1 mark) OR (ii)(b) What is length of diagonal PR? (1 mark)
(iii) If S divides CA in ratio K:1, find K. (2 marks) Made with ❤ by Sumeet Sahu | Unique Study Point Contact: Unique Study Point, Amitesh Nagar, Indore, MP | 8103405051 Visit: uniquestudyonline.com

Answer: (i) R(200, 400), S(–200, 400)
(ii)(a) Area = 400² = 160,000 sq units
OR (ii)(b) PR = 400√2 units
(iii) K = 1