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Class XI Chapter 10 –

Class XI Chapter 10 – Straight Lines Maths ______________________________________________________________________________ 2 ab p 2 2 a b p a b a b 2 2 2 2 2 2 2 2 a b 1 2 2 2 a b p 1 1 1 2 2 2 p a b Hence, we showed that 1 1 1 2 2 2 p a b Miscellaneous Exercise Question 1: Find the value of k for which the line (k – 3)x-(4 - (a) Parallel to x-axis, (b) Parallel to y-axis, (c) Passing through the origin. Solution 1: The given equation of line is 2 2 (k – 3)x-(4 - k )y + k - 7k + 6 = 0 …(1) (a) If the given line is parallel to the x-axis , then Slope of the given line = Slope of the x-axis The given line can be written as 2 2 (4-k )y=(k-3)x+ k 7k 6 0 2 (k-3) k 7k 6 x 2 2 2 k )y + y , which is of the form y = mx + c. (4-k ) (4-k ) Slope of the given line = (4-k 2 ) Slope of the x-axis =0 (k-3) 0 2 (4-k ) (k-3) k 3 0 k 3 Thus, the given line is parallel to x-axis , then the value of k is 3. 2 k - 7k + 6 = 0 is Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

Class XI Chapter 10 – Straight Lines Maths ______________________________________________________________________________ (b) If the given line is parallel to the y-axis, it is vertical. Hence, its slope will be undefined. (k-3) The slope of the given line is 2 (4-k ) (k-3) Now, 2 (4-k ) is undefined at 2 k = 4 2 k = 4 k = 2 Thus, if the given line is parallel to the y-axis, then the value of k is 2. (c)If the given line is passing through the origin, then point (0, 0) satisfies the given equation of line. 2 2 k3 0 (4-k ) 0 k 7k6 0 2 k 7k 6 0 2 k 6k k6 0 k k 6 1 0 k 1or 6 Thus, if the given line is passing through the origin, then the value of k is either 1 or 6. Question 2: Find the values of and p, if the equation x cos + y sin = p is the normal form of the line 3x y2 0 Solution 2: The equation of the given line is 3x y2 0 This equation can be reduced as 3x y2 0 3x y 2 2 2 On dividing both sides by 3 1 2 , we obtain 3 1 2 x y 2 2 2 3 1 x y1 ... 1 2 2 On comparing equation (1) to x cos + y sin = p, we obtain Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

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    Class XI Chapter 2 - Relations and

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    Class XI Chapter 3 - Trigonometric

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    Class XI Chapter 4 - Principle of M

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    Class XI Chapter 6 - Linear Inequal

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    Class XI Chapter 9 - Sequences and

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    Class XI Chapter 12 - Introduction

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    Class XI Chapter 13 - Limits and De

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    Class XI Chapter 14 - Mathematical

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    Class XI Chapter 15 - Statistics Ma

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