The vertices of a parallelogram in order are A (1, 2), B (4, y), C (x, 6) and D (3, 5). Then (x, y) is

(a) (6, 3)   (b) (3, 6)   (c) (5, 6)   (d) (1, 4)

 

This question is inspired from Ex 7.2, 6 (NCERT) - Chapter 7 Class 10 - Coordinate Geometry

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Question 29 The vertices of a parallelogram in order are A (1, 2), B (4, y), C (x, 6) and D (3, 5). Then (x, y) is (a) (6, 3) (b) (3, 6) (c) (5, 6) (d) (1, 4) Given A(1, 2), B(4, y), C(x, 6), D(3, 5) We know that Diagonals of parallelogram bisect each other So, O is the midβˆ’point of AC & BD Finding midβˆ’point of AC Coordinates of O = ((1 + π‘₯)/2,(2 + 6)/2) = ((1 + π‘₯)/2,8/2) = ((𝟏 + 𝒙)/𝟐,πŸ’) Finding midβˆ’point of BD Coordinates of O = ((3 + 4)/2,(5 + 𝑦)/2) = (πŸ•/𝟐,(5 + 𝑦)/2) Now, ((𝟏 + 𝒙)/𝟐,πŸ’) = (πŸ•/𝟐,(5 + 𝑦)/2) Comparing x-coordinate (1 + π‘₯)/2 = 7/2 (1 + x) = 7 x = 7 – 1 x = 6 Comparing x-coordinate 4 = (𝑦 + 5)/2 4 Γ— 2 = y + 5 8 – 5 = y y = 3 ∴ (x, y) = (6, 3) So, the correct answer is (a)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.