Examples

Chapter 7 Class 10 Coordinate Geometry
Serial order wise

### Transcript

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) Given that point A is equidistant from B & C Hence, AB = AC Finding AB & AC using distance formula Finding AB x1 = x, y1 = y x2 = 7, y2 = 1 AB = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √((7 −𝑥)2+(1 −𝑦)2) Similarly, finding AC x1 = x, y1 = y x2 = 3, y2 = 5 AC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 3 −𝑥)2+(5 −𝑦)2) We know that AB = AC √(( 7 −𝑥)2+(1 −𝑦)2) = √(( 3 −𝑥)2+(5 −𝑦)2) Squaring both sides 〖(√(( 7 −𝑥)2+(1 −𝑦)2) ")2 = (" √(( 3 −𝑥)2+(5 −𝑦)2))〗^2 (7 − x)2 + (1 − y)2 = (3 − x) 2 + (5 − y) 2 72 + x2 – 2 (7) (x) + 12 + y2 – 2y = (3) 2 + x2 − 2 (3)x + 52 + y2 – 2 (5)y 49 + x2 – 14x + 1 + y2 –2y = 9 + x2 − 6x + 25 + y2 – 10y (x2 − x2) – 14x + 6x + y2 − y2 −2y + 10y + 9 – 49 – 1 + 25 = 0 −8x + 8y + 16 = 0 8(−x + y + 2) = 0 −x + y + 2 = 0/8 −x + y + 2 = 0 y + 2 = x x = y + 2 x − y = 2 Since we only have to give relation between x & y So, answer is x − y = 2

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.