Misc 16 - Find direction in which a straight line must be drawn

Misc 16 - Chapter 10 Class 11 Straight Lines - Part 2
Misc 16 - Chapter 10 Class 11 Straight Lines - Part 3 Misc 16 - Chapter 10 Class 11 Straight Lines - Part 4 Misc 16 - Chapter 10 Class 11 Straight Lines - Part 5 Misc 16 - Chapter 10 Class 11 Straight Lines - Part 6

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Misc 15 Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point. Let line AB be x + y = 4 & point P be (–1, 2) Q is the point of intersection of line through P(-1, 2) to AB & PQ = 3 units We need to find the equation of line PQ First we will find the co-ordinates of point Q Since point Q(h,k) lies on line AB It will satisfy its equation x + y = 4 h + k = 4 h = 4 – k Also, Given PQ = 3 √((𝑥_2 − 𝑥_1 )^2 + (𝑦_2 − 𝑦_1 )^2 ) = 3 Putting values √((ℎ−(−1))^2 + (𝑘−2)^2 ) = 3 √((ℎ+1)^2 + (𝑘−2)^2 ) = 3 Squaring both sides (√((ℎ+1)^2 + (𝑘−2)^2 ))^2 = 32 (ℎ+1)^2 + (𝑘−2)^2 = 9 h2 + 12 + 2h + k2 + 22 – 4k = 9 h2 + 1 + 2h + k2 + 4 – 4k = 9 h2 + 2h + k2 – 4k = 9 – 1 – 4 h2 + 2h + k2 – 4k = 4 Putting h = 4 – k from (1) (4 – k)2 + 2(4 – k) + k2 – 4k = 4 42 + k2 – 8k + 8 – 2k + k2 – 4k = 4 16 + k2 – 8k + 8 – 2k + k2 – 4k = 4 k2 + k2 – 8k – 2k – 4k + 16 + 8 – 4 = 0 2k2 – 14k + 20 = 0 2(k2 – 7k + 10) = 0 k2 – 7k + 10 = 0 k2 – 5k – 2k + 10 = 0 k(k – 5) – 2(k – 5) = 0 (k – 2)(k – 5)= 0 So, k = 2 & k = 5 Now, finding equation of line PQ We know that equation of line joining two point (x1, y1) & (x2, y2) is (y – y1) = (𝑦2 − 𝑦1)/(𝑥2 − 𝑥1) (x – x1) So, line is parallel to x axis or y -axis

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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.