Question 16 (OR 2 nd  question)

Find the value of k for which the points (3k – 1, k – 2), (k, k – 7) and (k – 1, –k – 2) are collinear.

Find value of k for which points (3k - 1, k - 2), (k, k - 7) and (k-1

Question 16 (Or 2nd) - CBSE Class 10 Sample Paper for 2019 Boards - Part 2
Question 16 (Or 2nd) - CBSE Class 10 Sample Paper for 2019 Boards - Part 3

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Transcript

Question 16 (OR 2nd question) Find the value of k for which the points (3k – 1, k – 2), (k, k – 7) and (k – 1, –k – 2) are collinear. Let points be A (3k – 1, k – 2), B (k, k – 7) and C (k – 1, –k – 2) If A, B, C are collinear, they will lie on the same line, i.e. they will not form triangle Therefore, Area of ∆ABC = 0 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] = 0 1 mark Here x1 = 3k – 1 , y1 = k – 2 x2 = k , y2 = k – 7 x3 = k – 1 , y3 = −k – 2 Putting values 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] = 0 x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0 (3k – 1)[k – 7 – (–k – 2)] + k[−k − 2 – (k – 2)] + (k – 1)[(k – 2) – (k – 7)] = 0 (3k – 1)[k – 7 + k + 2] + k[−k − 2 – k + 2] + (k – 1)[k – 2 – k + 7] = 0 (3k – 1)(2k – 5) + k[−2k] + (k – 1)[–2 + 7] = 0 3k(2k – 5) – 1(2k – 5) – 2k2 + (k – 1)5 = 0 6k2 – 15k – 2k + 5 – 2k2 + 5k – 5 = 0 6k2 – 2k2 – 15k – 2k + 5k + 5 – 5 = 0 4k2 – 12k = 0 4k(k – 3) = 0 1 mark So, k = 0 and k = 3 1 mark

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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.