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Ex 7.1, 6 Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0) Let the points be A(−1, −2), B(1, 0), C(−1, 2), D(−3, 0) We find the distances AB, BC, CD & AD By using distance formula Finding AB AB = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 1 −(−1))2+(0 −(−2))2) = √((1+1)2+(0+2)2) = √((2)2+(2)2) = √(2(2)2) = 2√2 Finding BC BC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( −1 −1)2+(2 −0)2) = √((−2)2+(2)2) = √((2)2+(2)2) = √(2(2)2) = 2√2 Finding BC BC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( −1 −1)2+(2 −0)2) = √((−2)2+(2)2) = √((2)2+(2)2) = √(2(2)2) = 2√2 Finding CD CD = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( −3 −(−1))2+(0 −2)2) = √((−3+1)2+(−2)2) = √((−2)2+(−2)2) = √((2)2+(2)2) = √(2(2)2) = 2√2 Finding CD CD = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( −3 −(−1))2+(0 −2)2) = √((−3+1)2+(−2)2) = √((−2)2+(−2)2) = √((2)2+(2)2) = √(2(2)2) = 2√2 Finding AD AD = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( −3 −(−1))2+(0 −(−2))2) = √((−3+1)2+(2)2) = √((−2)2+(2)2) = √((2)2+(2)2) = √(2(2)2) = 2√2 Since, AB = BC = CD = AD = 2√2 All sides of quadrilateral are equal, ∴ It can be a rhombus or a square. To check if it is a square, we find diagonals AC and BD Finding AC AC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( −1 −(−1))2+(2 −(−2))2) = √((−1+1)2+(2+ 2)2) = √((0)2+(4)2) = √((4)2) = 4 Now, Finding BD BD = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 1 −(−3))2+(0 −0)2) = √(( 1+3)2+(0)2) = √((4)2+(0)2) = √((4)2) = 4 Since, AB = BC = CD = AD = 2√2 All sides of quadrilateral are equal, & AC = BD = 4 Hence, both diagonals are equal Since, all sides and both diagonals are equal ABCD is a square

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.