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Ex 7.1, 10 Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4). Let the points be A(x , y) , B(3, 6) , C(−3, 4) 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 = 3, y2 = 6 AB = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 3 −𝑥)2+(6 −𝑦)2) Similarly, Finding AC x1 = x, y1 = y x2 = −3, y2 = 4 AC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( −3 −𝑥)2+(4 −𝑦)2) We know AB = AC √(( 3 −𝑥)2+(6 −𝑦)2) = √(( −3 −𝑥)2+(4 −𝑦)2) Squaring both sides (√(( 3 −𝑥)2+(6 −𝑦)2) ")2 = (" √(( −3 −𝑥)2+(4 −𝑦)2))^2 (3 − x)2 + (6 − y)2 = (−3 − x) 2 + (4 − y) 2 32 + x2 – 6x + 62 + y2 – 12y = (−3) 2 + x2 + (–2) × (−3) × x + 42 + y2 – 8y 9 + x2 – 6x + 36 + y2 – 12y = 9 + x2 + 6x + 16 + y2 – 8y (x2 − x2) – 6x – 6x + y2 − y2 −12y + 8y + 9 – 9 – 16 + 36 = 0 −12x – 4y + 20 = 0 −4(3x + y – 5) = 0 3x + y – 5 = 0/5 3x + y – 5 = 0 Note that we only have to give relation between x & y and not solve it So, answer is 3x + y – 5 = 0

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.