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Ex 7.1, 1 - Find distance between: (i) (2, 3), (4, 1) - Distance Formula

  1. Chapter 7 Class 10 Coordinate Geometry
  2. Serial order wise
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Ex 7.1 , 1 Find the distance between the following pairs of points : (2, 3), (4, 1) Let the two points be P(2, 3) & Q(4, 1) We need to find distance PQ PQ = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) Here, x1 = x coordinate of P = 2 y1 = y coordinate of P = 3 x2 = x coordinate of Q = 4 y2 = y coordinate of Q = 1 Putting values PQ = √((4 −2 )2+(1 −3)2) = √((2 )2+(−2)2) = √((2 ×2)+(−2 ×−2) ) = √(4+4) = √8 = √(4(2)) = √(4 ×(2)) = √4 × √2 = 2 × √2 = 2√2 Ex 7.1 , 1 Find the distance between the following pairs of points : (−5, 7), (−1, 3) Let the two points be A(−5, 7) & B(−1, 3) We need to find distance AB AB = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) Here, x1 = x coordinate of A = −5 y1 = y coordinate of A = 7 x2 = x coordinate of B = −1 y2 = y coordinate of B = 3 Putting values AB = √((−1 −(−5 ))2+(3 −7)2) = √((−1+5 )2+(−4)2) = √((4)2+(−4)2) = √(16+16) = √32 = √(16(2)) = √(16 ×(2)) = √16 × √2 = 4 × √2 = 4√2 Ex 7.1 , 1 Find the distance between the following pairs of points : (a, b), (−a, −b) Let the two points be A(a, b) & B(−a, −b) We need to find distance AB AB = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) Here, x1 = x coordinate of A = a y1 = y coordinate of A = b x2 = x coordinate of B = −a y2 = y coordinate of B = −b Putting values AB = √((𝑎 −(−𝑎 ))2+(𝑏 −(−𝑏))2) = √((𝑎+𝑎)2+(𝑏+𝑏)2) = √((2𝑎)2+(2𝑏)2) = √((2𝑎)2+(2𝑏)2) = √(4𝑎2+4𝑏2) = √(4×(𝑎2+𝑏2)) = √4 × √(𝑎2+𝑏2) = 2 × √(𝑎2+𝑏2) = 2√(𝑎2+𝑏2)

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