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


CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [MCQ]
CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [MCQ]
Last updated at Dec. 16, 2024 by Teachoo
This question is inspired from Ex 7.2, 6 (NCERT) - Chapter 7 Class 10 - Coordinate Geometry
Transcript
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 midβpoint of AC & BD Finding midβpoint of AC Coordinates of O = ((1 + π₯)/2,(2 + 6)/2) = ((1 + π₯)/2,8/2) = ((π + π)/π,π) Finding midβpoint of BD Coordinates of O = ((3 + 4)/2,(5 + π¦)/2) = (π/π,(5 + π¦)/2) Now, ((π + π)/π,π) = (π/π,(5 + π¦)/2) Comparing x-coordinate (1 + π₯)/2 = 7/2 (1 + x) = 7 x = 7 β 1 x = 6 Comparing x-coordinate 4 = (π¦ + 5)/2 4 Γ 2 = y + 5 8 β 5 = y y = 3 β΄ (x, y) = (6, 3) So, the correct answer is (a)