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Example 4 - Find a relation between x and y such that - Examples

  1. Chapter 7 Class 10 Coordinate Geometry
  2. Serial order wise
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Example 4 Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5). Let the points be A(x, y) , B(7, 1) , C(3, 5) According to question, point A is equidistant from B & C Hence AB = AC So, we will find AB & AC using distance formula Finding AB x1 = x , y1 = y x2 = 7 , y2 = 1 AB = √((π‘₯2 βˆ’π‘₯1)2+(𝑦2 βˆ’π‘¦1)2) = √((7 βˆ’π‘₯)2+(1 βˆ’π‘¦)2) Similarly, finding AC x1 = x , y1 = y x2 = 3 , y2 = 5 AC = √((π‘₯2 βˆ’π‘₯1)2+(𝑦2 βˆ’π‘¦1)2) = √(( 3 βˆ’π‘₯)2+(5 βˆ’π‘¦)2) We know that AB = AC √(( 7 βˆ’π‘₯)2+(1 βˆ’π‘¦)2) = √(( 3 βˆ’π‘₯)2+(5 βˆ’π‘¦)2) Squaring both sides (√(( 7 βˆ’π‘₯)2+(1 βˆ’π‘¦)2))2 = (√(( 3 βˆ’π‘₯)2+(5 βˆ’π‘¦)2) ) 2 (7 βˆ’ x)2 + (1 βˆ’ y)2 = (3 βˆ’ x) 2 + (5 βˆ’ y) 2 72 + x2 – 2 (7) (x) + 12 + y2 – 2y = (3) 2 + x2 βˆ’ 2 (3)x + 52 + y2 – 2 (5)y 49 + x2 – 14x + 1 + y2 –2y = 9 + x2 βˆ’ 6x + 25 + y2 – 10y (x2 βˆ’ x2) – 14x + 6x + y2 βˆ’ y2 βˆ’2y + 10y + 9 – 49 – 1 + 25 = 0 βˆ’8x + 8y + 16 = 0 8(βˆ’x + y + 2) = 0 βˆ’x + y + 2 = 0/8 βˆ’x + y + 2 = 0 y + 2 = x x = y + 2 Note that we only have to give relation between x & y and not solve it So, answer is x = y + 2

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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