Class 10 Mathematics: Coordinate Geometry important questions

Coordinate Geometry is a chapter of Class 10 Mathematics for the Andhra Pradesh State Board. These 11 questions are marked Very Important, Important or Practice, and each answer has been checked by our team. Use them to revise, not as a prediction of what will come in the exam.

1. Practice · 1 marks

   1. Find the midpoint of the segment joining (0, 0) and (4, 6).

   ((0 + 4)/2, (0 + 6)/2) = (2, 3).

2. Important · 2 marks

   2. Find the midpoint of the segment joining (cos 0°, 0) and (0, sin 90°).

   cos 0° = 1 and sin 90° = 1, so the points are (1, 0) and (0, 1). Midpoint = ((1 + 0)/2, (0 + 1)/2) = (1/2, 1/2).

3. Very Important · 8 marks

   3. Find the area of a rhombus whose vertices, taken in order, are (−4, −7), (−1, 2), (8, 5) and (5, −4).

   Let A(−4, −7), B(−1, 2), C(8, 5), D(5, −4). Diagonal AC = √[(8 + 4)² + (5 + 7)²] = √(144 + 144) = 12√2. Diagonal BD = √[(5 + 1)² + (−4 − 2)²] = √(36 + 36) = 6√2. Area of a rhombus = ½ × d₁ × d₂ = ½ × 12√2 × 6√2 = ½ × 144 = 72 square units.

4. Important · 2 marks

   4. Write the formula for the distance between i) the two points (x₁, y₁) and (x₂, y₂) ii) the origin and a point (x, y)

   i) d = √[(x₂ − x₁)² + (y₂ − y₁)²] ii) d = √(x² + y²)

5. Very Important · 8 marks

   5. Two students claim to have found the points of trisection of the segment joining A(4, −1) and B(−2, −3). Student A: (0, 5/3) and (0, 7/3) Student B: (2, −5/3) and (0, −7/3) Who is correct? Justify.

   The trisection points divide AB in the ratios 1 : 2 and 2 : 1. Ratio 1 : 2: ((1 × −2 + 2 × 4)/3, (1 × −3 + 2 × −1)/3) = (6/3, −5/3) = (2, −5/3). Ratio 2 : 1: ((2 × −2 + 1 × 4)/3, (2 × −3 + 1 × −1)/3) = (0/3, −7/3) = (0, −7/3). Student B is correct.

6. Important · 2 marks

   6. Find the coordinates of the midpoint of the segment joining (−2, −2) and (2, −4).

   Midpoint = ((−2 + 2)/2, (−2 + (−4))/2) = (0, −3).

7. Very Important · 8 marks

   7. Do the points (3, 2), (−2, −3) and (2, 3) form a triangle? If so, name the type of triangle.

   Let A(3, 2), B(−2, −3), C(2, 3). AB = √[(−5)² + (−5)²] = √50, BC = √[4² + 6²] = √52, CA = √[1² + 1²] = √2. The sum of any two sides is greater than the third (e.g. √50 + √2 ≈ 8.49 > √52 ≈ 7.21), so the points are not collinear and form a triangle. AB² + CA² = 50 + 2 = 52 = BC², so it is a right-angled triangle, with the right angle at A.

8. Important · 2 marks

   8. The vertices of a triangle are A(5, 1), B(1, 5) and C(−3, −1). Find the length of the median AD.

   D is the midpoint of BC: D = ((1 + (−3))/2, (5 + (−1))/2) = (−1, 2). AD = √[(5 − (−1))² + (1 − 2)²] = √(36 + 1) = √37 units.

9. Very Important · 8 marks

   9. The distance between the points P(2, −3) and Q(10, y) is 10 units. Jyothi says y has only one value, but Preethi disagrees. Resolve their conflict with a logical conclusion.

   PQ² = (10 − 2)² + (y + 3)² = 100 64 + (y + 3)² = 100 (y + 3)² = 36, so y + 3 = 6 or y + 3 = −6. y = 3 or y = −9. There are two values, so Preethi is right. Both Q(10, 3) and Q(10, −9) are 10 units from P.

10. Important · 2 marks

    10. A point lies 4 units to the left of the y-axis and 3 units below the x-axis. Many students plotted it in the first quadrant. Write the correct coordinates and name the quadrant.

    Left of the y-axis means x is negative, and below the x-axis means y is negative. The point is (−4, −3), in the third quadrant.

11. Very Important · 8 marks

    11. Four friends sit at A(3, 4), B(6, 7), C(9, 4) and D(6, 1). Champa says ABCD is a square; Chameli disagrees. Using the distance formula, find who is correct.

    AB = √[(6 − 3)² + (7 − 4)²] = √18 BC = √[(9 − 6)² + (4 − 7)²] = √18 CD = √[(6 − 9)² + (1 − 4)²] = √18 DA = √[(3 − 6)² + (4 − 1)²] = √18 Diagonals: AC = √[(9 − 3)² + 0²] = 6 and BD = √[0² + (1 − 7)²] = 6. All four sides are equal and the diagonals are equal, so ABCD is a square. Champa is correct.