Misc 24 - A person standing at junction of two straight paths

Misc 24 - Chapter 10 Class 11 Straight Lines - Part 2
Misc 24 - Chapter 10 Class 11 Straight Lines - Part 3 Misc 24 - Chapter 10 Class 11 Straight Lines - Part 4 Misc 24 - Chapter 10 Class 11 Straight Lines - Part 5 Misc 24 - Chapter 10 Class 11 Straight Lines - Part 6 Misc 24 - Chapter 10 Class 11 Straight Lines - Part 7 Misc 24 - Chapter 10 Class 11 Straight Lines - Part 8 Misc 24 - Chapter 10 Class 11 Straight Lines - Part 9

 

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Transcript

Misc 23 A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow. Let equation of lines OA : 2x − 3y + 4 = 0 OB : 3x + 4y – 5 = 0 AB : 6x – 7y + 8 = 0 The two paths cross at point O ∴ The person is standing at point O From point O, if he has to reach line AB in least time, Least distance from point O to line AB will be perpendicular distance, Let OM ⊥ AB We need to find equation of line OM First we find coordinates of point O Point O is the intersection of OA & OB From (1) 2x – 3y + 4 = 0 2x = 3y – 4 x = (3𝑦 − 4)/2 x = 3/2 y – 4/2 x = 3/2 y – 2 Putting value of x in (2) 3x + 4y – 5 = 0 3(3/2 𝑦−2) + 4y − 5 = 0 9/2y – 6 + 4y − 5 = 0 9/2y + 4y – 11 = 0 (9𝑦 + 8𝑦 )/2 = 11 9y + 8y = 22 17y = 22 y = 22/17 Putting value of y in (1) 2x – 3y + 4 = 0 2x – 3(22/17) + 4 = 0 2x – 66/17 + 4 = 0 2x = 66/17 – 4 2x = (66 − 4(17))/17 2x = (66 − 68)/17 2x = (−2)/17 x = (−2)/(17 × 2) x = (−1)/17 Thus, the coordinate of point O is O ((−𝟏)/𝟏𝟕, 𝟐𝟐/𝟏𝟕) Now, line OM is perpendicular to line AB When two lines are perpendicular then product of their slopes are equal to − 1 Thus, Slope of OM × Slope of AB = − 1 Slope of OM = ( − 1)/(𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝐴𝐵) Finding slope of AB Equation of AB is 6x − 7y + 8 = 0 6x + 8 = 7y 7y = 6x + 8 y = (6𝑥 + 8)/7 y = (6/7)x + 8/7 The above equation is of the form y = mx + c Where m is slope of line Slope of AB = 6/7 Slope of OM = ( −1)/(𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝐴𝐵) = ( −1)/(6/7) = (−7)/6 Finding equation of OM We know that equation of line passing through (x1, y1) having slope m is Equation of line OM passing through O((−1)/17, 22/17) having slope (−7)/6 is Equation of path OM is "y "−22/17 = (−7)/6 ("x" −((−1)/17)) "y "−22/17 = (−7)/6 ("x" +1/17) "y "−22/17 = (−7)/6 𝑥−7/6 × 1/17 "y "−22/17 = (−7)/6 𝑥−7/(6 × 17) "y "+ 7/6 x = 22/17 – 7/(6 × 17) (6𝑦 + 7𝑥)/6 = (22(6) − 7)/(6 × 17) (6𝑦 + 7𝑥)/6 = (132 −7)/(6 × 17) (6𝑦 + 7𝑥)/6 = 125/(6 × 17) 6y + 7x = 125/17 17(6y + 7x) = 125 17(6y) + 17(7x) = 125 102y + 119x = 125 Hence the required equation is 119x + 102y = 125

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