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Misc 1 - Find values of k for (k - 3) x - (4 - k2)y + k2

Misc 1 - Chapter 10 Class 11 Straight Lines - Part 2
Misc 1 - Chapter 10 Class 11 Straight Lines - Part 3 Misc 1 - Chapter 10 Class 11 Straight Lines - Part 4 Misc 1 - Chapter 10 Class 11 Straight Lines - Part 5 Misc 1 - Chapter 10 Class 11 Straight Lines - Part 6

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Misc 1 Find the values of k for which the line (k – 3) x – (4 – k2)y + k2 – 7k + 6 = 0 is Parallel to the x-axis, Any line parallel to x-axis is of the form y = p where p is constant So, there is no x term Since Line (k – 3) x – (4 – k2) y + k2 – 7k + 6 = 0 is parallel to x-axis Hence, (k − 3)x = 0 k – 3 = 0/𝑥 k – 3 = 0 k = 3 Misc 1 Find the values of k for which the line (k – 3) x – (4 – k2) y + k2 – 7k + 6 = 0 is (b) Parallel to the y-axis, Any line parallel to y-axis is of the form x = p where p is constant So, there is no y term Since line (k – 3) x – (4 – k2) y + k2 – 7k + 6 = 0 is parallel to y-axis Hence, –(4 – k2) y = 0 −(4 − k2) = 0/𝑦 –4 + k2 = 0 k2 = 4 k = ± √4 k = ± 2 Hence k = 2 or −2 Misc 1 Find the values of k for which the line (k – 3) x – (4 – k2) y + k2 – 7k + 6 = 0 is (c) Passing through origin If the line passing through the origin i.e. (0, 0) will satisfy the equation of line Putting x = 0 & y = 0 in equation (k − 3)x − (4 − k2)y + k2 − 7k + 6 = 0 (k − 3)0 − (4 − k2)0 + k2 − 7k + 6 = 0 k2 − 7k + 6 = 0 k2 − 6k − k + 6 = 0 k(k − 6) − 1(k − 6) = 0 k(k − 6) − 1(k − 6) = 0 (k − 1)(k − 6) = 0 So, k = 1 or k = 6

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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.