Example 2 - Chapter 7 Class 10 Coordinate Geometry

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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 = (( 2 1)2+( 2 1)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 = (( 2 1)2+( 2 1)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 = (( 2 1)2+( 2 1)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 = (( 2 1)2+( 2 1)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 = (( 2 1)2+( 2 1)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 = (( 2 1)2+( 2 1)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