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Ex 10.1, 4 - Find a point on x-axis, which is equidistant - Ex 10.1

  1. Chapter 10 Class 11 Straight Lines
  2. Serial order wise
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Ex10.1, 4 Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4). Let the given points be A(7, 6) & B (3, 4) Let C be a point on the x-axis Coordinates of C = C(x, 0) Given that point C is equidistant from the points A & B Hence, Distance AC = Distance BC We know that distance between two points (x1 , y1)&(x2 , y2)is D = √((𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2) Distance between A(7, 6) & C(x, 0) AC = √((𝑥 − 7)^2 + (0 − 6)2) = √((𝑥 − 7)^2 + 36) Distance between B(3, 4) & C(4, 0) BC = √((𝑥 − 3)^2 + (0 − 4)2) = √((𝑥 − 3)^2 + 16) Since, AC = BC √((𝑥 − 7)^2 + 36) = √((𝑥 − 3)^2 + 16) Squaring both sides (√((𝑥 − 7)^2 + 36))^2 = (√((𝑥 − 3)^2 + 16))^2 (x – 7)2 + 36 = (x – 3)2 + 16 (x – 7)2 – (x – 3)2 = 16 – 36 x2 + 49 – 14x – (x2 + 9 – 6x) = –20 x2 + 49 – 14x – x2 – 9 + 6x = –20 0 – 8x + 40 = -20 –8x = – 20 – 40 –8x = -60 x = ( − 60)/( − 8) x = 15/2 Thus, Required point = C(x ,0) = (15/2, 0)

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