# Example 2 - Chapter 7 Class 10 Coordinate Geometry (Term 1)

Last updated at Aug. 16, 2021 by Teachoo

Examples

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Example 2 Important You are here

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Example 7 Important

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Example 10 Important

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Example 13 Important Deleted for CBSE Board 2022 Exams

Example 14 Important Deleted for CBSE Board 2022 Exams

Example 15 Deleted for CBSE Board 2022 Exams

Last updated at Aug. 16, 2021 by Teachoo

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?