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Example 25 - Show that path of a moving point such that distance

Example 25 - Chapter 10 Class 11 Straight Lines - Part 2
Example 25 - Chapter 10 Class 11 Straight Lines - Part 3 Example 25 - Chapter 10 Class 11 Straight Lines - Part 4

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Example 25 Show that the path of a moving point such that its distances from two lines 3x – 2y = 5 and 3x + 2y = 5 are equal is a straight line. Given lines are 3x – 2y = 5 i.e. 3x – 2y – 5 = 0 & 3x + 2y = 5 i.e. 3x + 2y – 5 = 0 Let point (h, k) be any point whose distance from line (1) & (2) equal We know that distance of a point (x1, y1) from line Ax + By + C = 0 is d = |〖𝐴𝑥〗_1 + 〖𝐵𝑦〗_1 + 𝐶|/√(𝐴^2 + 𝐵^2 ) Here distance of a point (h, k) from line 3x – 2y – 5 = 0 is d = |〖𝐴𝑥〗_1 + 〖𝐵𝑦〗_1 + 𝑐|/√(𝐴^2 + 𝐵^2 ) Here x1 = h, y1 = k & A = 3, B = −2 , C = −5 Putting values d = |3(ℎ) + (−2)(𝑘) + (5)|/√((3)^2 + (−2)^2 ) d = |3ℎ − 2𝑘 + 5|/√13 Distance of point (h, k) from a line 3x + 2y – 5 = 0 is d = |〖𝐴𝑥〗_1 + 〖𝐵𝑦〗_1 + 𝑐|/√(𝐴^2 + 𝐵^2 ) Here x1 = h, y1 = k & A = 3, B = 2, C = –5 Putting values d = |3(ℎ) + 2(𝑘) − 5|/√(〖(3)〗^2 + 〖(2)〗^2 ) d = |3ℎ + 2𝑘 − 5|/√13 Since both distance are equal |3ℎ − 2𝑘 − 5|/√13 = |3ℎ + 2𝑘 − 5|/√13 |3ℎ−2𝑘−5| = |3ℎ+2𝑘−5| 3h – 2k – 5 = ± (3h + 2k – 5) So, 3h – 2k – 5 = 3h + 2k – 5 or 3h – 2k – 5 = – ( 3h + 2k – 5) Solving (A) 3h – 2k – 5 = 3h + 2k – 5 3h – 2k – 5 – 3h – 2k + 5 = 0 3h – 3h – 2k – 2k – 5 + 5 = 0 –4k = 0 k = 0 Thus point (h, k) satisfies equation y = 0 Solving (B) 3h – 2k – 5 = –(3h + 2k – 5) 3h – 2k – 5 + 3h + 2k – 5 = 0 6h = 10 h = 10/6 = 5/3 Thus the point (h, k) satisfies equation x = 5/3 ∴ The point (h, k) satisfies the equation of line y = 0 or x = 5/3 which represents straight line Hence, path of the point equidistant from the line (1) & (2) is straight line

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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.