Question 13 - End-of-Chapter Exercises - Chapter 1 Class 9 - Orienting Yourself: The Use of Coordinates (Ganita

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Question 13 The midpoints of the sides of triangle ABC are the points D, E, and F. Given that the coordinates of D, E, and F are (5, 1), (6, 5), and (0, 3), respectively, find the coordinates of A, B and C. We draw a diagram Let A (x1, y1) , B (x2, y2), C (x3, y3) Now, D is mid-point of BC E is mid-point of AB F is mid-point of AC We first solve point F since it’s x-coordinate is 0 F is mid-point of AC Coordinates of F = ((𝒙_𝟏 + 𝒙_𝟑)/𝟐,(𝒚_𝟏 + 𝒚_𝟑)/𝟐) (0, 3) = ((𝒙_𝟏 + 𝒙_𝟑)/𝟐,(𝒚_𝟏 + 𝒚_𝟑)/𝟐) Comparing x and y coordinates 𝟎=(𝒙_𝟏 + 𝒙_𝟑)/𝟐 0 × 2 = 𝑥_1 + 𝑥_3 0 = 𝑥_1 + 𝑥_3 𝒙_𝟏 + 𝒙_𝟑 = 0 …(1) 𝟑=(𝒚_𝟏 + 𝒚_𝟑)/𝟐 3 × 2 = 𝑦_1 + 𝑦_3 6 = 𝑦_1 + 𝑦_3 𝒚_𝟏 + 𝒚_𝟑 = 6 …(2) E is mid-point of AB Coordinates of E = ((𝒙_𝟏 + 𝒙_𝟐)/𝟐,(𝒚_𝟏 + 𝒚_𝟐)/𝟐) (6, 5) = ((𝒙_𝟏 + 𝒙_𝟐)/𝟐,(𝒚_𝟏 + 𝒚_𝟐)/𝟐) Comparing x and y coordinates 𝟔=(𝒙_𝟏 + 𝒙_𝟐)/𝟐 6 × 2 = 𝑥_1 + 𝑥_2 12 = 𝑥_1 + 𝑥_2 𝒙_𝟏 + 𝒙_𝟐 = 12 …(3) 𝟓=(𝒚_𝟏 + 𝒚_𝟐)/𝟐 5 × 2 = 𝑦_1 + 𝑦_2 10 = 𝑦_1 + 𝑦_2 𝒚_𝟏 + 𝒚_𝟐 = 10 …(4) D is mid-point of BC Coordinates of D = ((𝒙_𝟐 + 𝒙_𝟑)/𝟐,(𝒚_𝟐 + 𝒚_𝟑)/𝟐) (5, 1) = ((𝒙_𝟐 + 𝒙_𝟑)/𝟐,(𝒚_𝟐 + 𝒚_𝟑)/𝟐) Comparing x and y coordinates 𝟓=(𝒙_𝟐 + 𝒙_𝟑)/𝟐 5 × 2 = 𝑥_2 + 𝑥_3 10 = 𝑥_2 + 𝑥_3 𝒙_𝟐 + 𝒙_𝟑 = 10 …(5) 𝟏=(𝒚_𝟐 + 𝒚_𝟑)/𝟐 1 × 2 = 𝑦_2 + 𝑦_3 2 = 𝑦_2 + 𝑦_3 𝒚_𝟐 + 𝒚_𝟑 = 2 …(6) Now, our equations are 𝑥_1 + 𝑥_3 = 0 …(1) 𝑥_1 + 𝑥_2 = 12 …(3) 𝑥_2 + 𝑥_3 = 10 …(5) Adding (1), (3) and (5) (𝑥_1 + 𝑥_3) + (𝑥_1 + 𝑥_2) + (𝑥_2 + 𝑥_3) = 0 + 12 + 10 2 × (𝒙_𝟏+ 𝒙_𝟐+ 𝒙_𝟑) = 22 (𝑥_1+ 𝑥_2+ 𝑥_3) = 22/2 𝒙_𝟏+ 𝒙_𝟐+ 𝒙_𝟑= 11 Now, we can write 𝑥_1+ 𝑥_2+ 𝑥_3= 11 𝒙_𝟏=𝟏𝟏−(𝒙_𝟐+ 𝒙_𝟑) From (1): 𝑥_1 + 𝑥_3 = 10 𝑥_1=11−10 𝒙_𝟏=𝟏 Similarly, 𝑥_1+ 𝑥_2+ 𝑥_3= 11 𝒙_𝟐=𝟏𝟏−(𝒙_𝟏+ 𝒙_𝟑) From (1): 𝑥_1 + 𝑥_3 = 0 𝑥_2=11−0 𝒙_𝟐=𝟏𝟏 Also, 𝑥_1+ 𝑥_2+ 𝑥_3= 11 𝒙_𝟑=𝟏𝟏−(𝒙_𝟏+ 𝒙_𝟐) From (3): 𝑥_1 + 𝑥_2 = 12 𝑥_3=11−12 𝒙_𝟑=−𝟏 Similarly 𝑦_1 + 𝑦_3 = 6 …(2) 𝑦_1 + 𝑦_2 = 10 …(4) 𝑦_2 + 𝑦_3 = 2 …(6) Adding (2), (4) and (6) (𝑦_1 + 𝑦_3) + (𝑦_1 + 𝑦_2) + (𝑦_2 + 𝑦_3) = 6 + 10 + 2 2 × (𝒚_𝟏+ 𝒚_𝟐+ 𝒚_𝟑) = 18 (𝑦_1+ 𝑦_2+ 𝑦_3) = 18/2 𝒚_𝟏+ 𝒚_𝟐+ 𝒚_𝟑= 9 Now, we can write 𝑦_1+ 𝑦_2+ 𝑦_3= 9 𝒚_𝟏=𝟗−(𝒚_𝟐+ 𝒚_𝟑) From (6): 𝑦_2 + 𝑦_3 = 2 𝑦_1=9−2 𝒚_𝟏=𝟕 Similarly, 𝑦_1+ 𝑦_2+ 𝑦_3= 9 𝒚_𝟐=𝟗−(𝒚_𝟏+ 𝒚_𝟑) From (2): 𝑦_1 + 𝑦_3 = 6 𝑦_2=9−6 𝒚_𝟐=𝟑 Also, 𝑦_1+ 𝑦_2+ 𝑦_3= 9 𝒚_𝟑=𝟗−(𝒚_𝟏+ 𝒚_𝟐) From (4): 𝑦_1 + 𝑦_2 = 10 𝑦_3=9−10 𝒚_𝟑=−𝟏 So, x1 = 1, y1 = 7 x2 = 11, y2 = 3 x3 = –1, y3 = –1 Thus, the corners of the triangle are A(1, 7), B(11, 3), and C(-1, -1)