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Example 2 Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square. Let the points be A(−1, −2), B(1, 0), C(−1, 2), D(−3, 0) To prove that ABCD is a square, We have to prove all sides equal, i.e. AB = BC = CD = AD & Diagonals equal i.e. AC = BD We find the distances AB, BC, CD & AD and AC & BD Finding AB x1 = 1 , y1 = 7 x2 = 4 , y2 = 2 AB = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √((4 −1)2+(2 −7)2) = √((3)2+(−5)2) = √((3)2+(5)2) = √(9+25) = √𝟑𝟒 Finding BC x1 = 4 , y1 = 2 x2 = −1 , y2 = −1 BC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( −1 −4)2+(−1 −2)2) = √((−5)2+(−3)2) = √((5)2+(3)2) = √(25+9) = √𝟑𝟒 Finding CD x1 = −1 , y1 = −1 x2 = −4 , y2 = 4 CD = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( −4 −(−1))2+(4 −(−1))2) = √((−4+1)2+(4+1)2) = √((−3)2+(5)2) = √((3)2+(5)2) = √(9+ 25) = √𝟑𝟒 Finding AD x1 = 1 , y1 = 7 x2 = −4 , y2 = 4 AD = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( −4 −1)2+(4 −7)2) = √((−5)2+(−3)2) = √((5)2+(3)2) = √(25+9) = √𝟑𝟒 Now, finding diagonals Finding AC x1 = 1 , y1 = 7 x2 = −1 , y2 = −1 AC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( −1 −1)2+(−1 −7)2) = √((−2)2+(−8)2) = √((2)2+(8)2) = √(4+64) = √𝟔𝟖 Now, finding diagonals Finding BD x1 = 4 , y1 = 2 x2 = −4 , y2 = 4 BD = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √((−4 −4)2+(4 −2)2) = √((−8)2+(2)2) = √((8)2+(2)2) = √(64 +4) = √𝟔𝟖 Since, AB = BC = CD = AD = √34 All sides of ABCD are equal, & AC = BD = √68 Hence, both diagonals are equal Since, all sides and both diagonals are equal Hence, ABCD is a square Question for you: Is there a better method?

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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.