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Misc 20 - If sum of perpendicular distances of variable point

Misc 20 - Chapter 10 Class 11 Straight Lines - Part 2
Misc 20 - Chapter 10 Class 11 Straight Lines - Part 3 Misc 20 - Chapter 10 Class 11 Straight Lines - Part 4 Misc 20 - Chapter 10 Class 11 Straight Lines - Part 5 Misc 20 - Chapter 10 Class 11 Straight Lines - Part 6

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Misc 20 If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line. Given lines are x + y – 5 = 0 & 3x – 2y + 7 = 0 We know that , Perpendicular distance of a point (x1, y1) from line Ax + By + C = 0 is d = |〖𝐴𝑥〗_1 + 〖𝐵𝑦〗_1 + 𝐶|/√(𝐴^2 + 𝐵^2 ) Let p1 be the perpendicular distance of point P(x, y) from line x + y – 5 = 0 Here x1 = x , y1 = y & A = 1 , B = 1 , C = − 5 p1 = |1(𝑥) + 1(𝑦) + (−5)|/√((1)^2 + (1)^2 ) = |𝑥 + 𝑦 − 5|/√(1 + 1) = |𝑥 + 𝑦 − 5|/√2 Let p2 be the perpendicular distance from point P(x, y) from line 3x – 2y + 7 = 0 is Here x1 = x , y1 = y & A = 3 , B = –2 , C = 7 p2 = |3(𝑥) − 2(𝑦) − (7)|/√((3)^2 + ( − 2)^2 ) = |3𝑥 − 2𝑦 − 7|/√(9 + 4) = |3𝑥 − 2𝑦 − 7|/√13 Given that Sum of perpendicular distance = 10 p1 + p2 = 10 |𝑥 + 𝑦 − 5|/√2 + |3𝑥 − 2𝑦 + 7|/√13 = 10 (√13 |𝑥 + 𝑦 − 5| + √2 |3𝑥 − 2𝑦 + 7|)/(√2 × √13) = 10 √13 |"(x" + "y" − "5)" | + √2 |"(3x – 2y" + "7)" | = 10 × √13 × √2 √13 |"(x" + "y" − "5)" | + √2 |"(3x – 2y" + "7)" | = 10 √26 We will get 4 cases by solving this equation 1. Taking (x + y – 5 ) & (3x – 2y + 7) positive 2. Taking (x + y – 5) positive & (3x – 2y + 7) negative 3. Taking (x + y – 5) negative & (3x – 2y + 7) positive 4. Taking both (x + y – 5) & (3x – 2y + 7) negative Now, solving separately Taking (x + y – 5 ) & (3x – 2y + 7) positive √13 "(x" + "y" − "5)" + √2 "(3x – 2y" + "7)" = 10 √26 √13x + √13y - 5√13 + 3√2x − 2√2 + 7√2 = 10√26 (√13 + 3√2)x + (√13 − 3√2)y - 5√13 + 7√2 - 10√26 = 0 Which is the equation of a straight line Taking (x + y – 5) positive & (3x – 2y + 7) negative √13(x + y – 5) + √2[ − (3x – 2y + 7)] = 10 √26 √13(x + y – 5) − √2(3x – 2y + 7) = 10 √26 √13x + √13y − 5√13 − 3√2x + 2√2y − 7√2 − 10√26 = 0 (√13 + 3√2)x + (√13 + 2√2)y − 5√13 − 7√2 − 10√26 = 0 Which is the equation of a straight line Taking (x + y – 5) negative & (3x – 2y + 7) positive −√13 (x + y – 5) + √2 (3x – 2y + 7) = 10 √26 −√13x + √13y + 5√13 + 3√2x − 2√2y + 7√2 − 10√26 = 0 (−√13+3√2)x + (− √13 − 2√2)y + 5√13 +7√2 − 10√26 = 0 Which is the equation of a straight line Taking √𝟏𝟑 (x + y – 5) & √𝟐 (3x – 2y + 7) both negative −√13(x + y − 5) − √2(3x − 2y + 7) = 10 √26 −√13x − √13y + 5√13 − 3√2x + 2√2y − 7√2 − 10√26 = 0 (−√13−3√2)x + (−√13+2√2)y + 5√13 − 7√2 − 10√26 = 0 Which is the equation of a straight line Thus, from all 4 cases point P(x, y) moves on a straight line Hence proved

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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.