Question 2 - Case Based Questions (MCQ) - Chapter 7 Class 10 Coordinate Geometry

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Question The class X students school in krishnagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot. Marking points on the image Niharika runs 1/4th of the distance AD on 2nd line So, Niharika’s x−cordinate = 2 Niharika’ y – cordinate = 1/4 × 100 = 25 ∴ Coordinates of Niharika = G (2, 25) Question 1 Taking A as origin, find the coordinates of P (a) (4, 6) (b) (6, 4) (c) (0, 6) (d) (4, 0) If A is the Origin Coordinates of point P = (4, 6) Coordinates of point Q = (3, 2) Coordinates of point R = (6, 5) So, the correct answer is (a) Question 2 What will be the coordinates of R, if C is the origin? (a) (8, 6) (b) (3, 10) (c) (10, 3) (d) (0, 6) If C is the Origin (and taking downward as positive) Coordinates of point P = (12, 2) Coordinates of point Q = (13, 6) Coordinates of point R = (10, 3) So, the correct answer is (c) Question 3 What will be the coordinates of Q, if C is the origin? (a) (6, 13) (b) (–6, 13) (c) (–13, 6) (d) (13, 6) Coordinates of point Q = (13, 6) So, the correct answer is (d) Question 4 Calculate the area of the triangles if A is the origin (a) 4.5 (b) 6 (c) 8 (d) 6.25 If A is the origin, We need to find Area of Δ PQR where P(4, 6) , Q(3, 2) and R(6, 5) Area of ∆PQR with vertices P (4, 6), Q (3, 2) & R (6, 5) Here, 𝒙_𝟏=4 , 〖 𝒚〗_𝟏=6 𝒙_𝟐=3 , 𝒚_𝟐=2 𝒙_𝟑=6 , 𝒚_𝟑=5 Now, ar (∆PQR) = 1/2 [𝑥_1 (𝑦_2−𝑦_3 )+ 𝑥_2 (𝑦_3−𝑦_1 )+𝑥_3 (𝑦_1−𝑦_2 )] = 𝟏/𝟐 [𝟒 (𝟐−𝟓)+𝟑 (𝟓−𝟔)+ 𝟔 (𝟔−𝟐)] = 1/2 [4 (−3)+3 (−1)+6 (4)] = 1/2 [−12−3+24] = 1/2 × 9 = 4.5 sq. units So, the correct answer is (a) Question 5 Calculate the area of the triangles if C is the origin (a) 8 (b) 5 (c) 6.25 (d) 4.5 If C is the origin, We need to find Area of Δ PQR where P(12, 2) , Q(13, 6) and R(10, 3) Area of ∆PQR with vertices P (12, 2), Q (13, 6) & R (10, 3) Here, 𝒙_𝟏=12 , 𝒚_𝟏=2 𝒙_𝟐=13 , 𝒚_𝟐=6 𝒙_𝟑=10 , 𝒚_𝟑=3 Now, ar (∆PQR) = 1/2 [𝑥_1 (𝑦_2−𝑦_3 )+ 𝑥_2 (𝑦_3−𝑦_1 )+𝑥_3 (𝑦_1−𝑦_2 )] = 𝟏/𝟐 [𝟏𝟐(𝟔−𝟑)+𝟏𝟑(𝟑−𝟐)+𝟏𝟎(𝟐−𝟔)] = 1/2 [12 × 3+13 × 1+10(−4)] = 1/2 [36+13−40] = 1/2 × 9 = 4.5 sq. units So, the correct answer is (d)