Example 3 - Figure shows arrangement of desks in a classroom - Examples

part 2 - Example 3 - Examples - Serial order wise - Chapter 7 Class 10 Coordinate Geometry
part 3 - Example 3 - Examples - Serial order wise - Chapter 7 Class 10 Coordinate Geometry part 4 - Example 3 - Examples - Serial order wise - Chapter 7 Class 10 Coordinate Geometry part 5 - Example 3 - Examples - Serial order wise - Chapter 7 Class 10 Coordinate Geometry

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Example 3 Given figure shows the arrangement of desks in a classroom. Ashima, Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6) respectively. Do you think they are seated in a line? Give reasons for your answer. If Ashima, Bharti & Camella are seated in a line, then points A, B & C must be collinear i.e. AB + BC = AC We Find AB , BC & AC using distance formula Finding AB x1 = 3 , y1 = 1 x2 = 6 , y2 = 4 AB = โˆš((๐‘ฅ2 โˆ’๐‘ฅ1)2+(๐‘ฆ2 โˆ’๐‘ฆ1)2) = โˆš((6 โˆ’3)2+(4 โˆ’1)2) = โˆš((3)2+(3)2) = โˆš(2(3)2) = (โˆš2)3 = 3โˆš๐Ÿ Finding BC x1 = 6 , y1 = 4 x2 = 8 , y2 = 6 BC = โˆš((๐‘ฅ2 โˆ’๐‘ฅ1)2+(๐‘ฆ2 โˆ’๐‘ฆ1)2) = โˆš((8 โˆ’6)2+(6 โˆ’4)2) = โˆš((2)2+(2)2) = โˆš(2(2)2) = (โˆš2)2 = 2โˆš๐Ÿ Finding AC x1 = 3, y1 = 1 x2 = 8, y2 = 6 AC = โˆš((๐‘ฅ2 โˆ’๐‘ฅ1)2+(๐‘ฆ2 โˆ’๐‘ฆ1)2) = โˆš((8 โˆ’3)2+(6 โˆ’1)2) = โˆš((5)2+(5)2) = โˆš(2(5)2) = (โˆš2)5 = 5โˆš๐Ÿ Hence AB = 3โˆš2, BC = 2โˆš2, AC = 5โˆš2 Now, AB + BC = 3โˆš2 + 2โˆš2 = 5โˆš2 = AC So, points A, B & C are collinear Hence, Ashima, Bharti and Camilla are seated in a line

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