Class XI Chapter 10 – Straight Lines Maths ______________________________________________________________________________ 3 cos = - 2 , sin = 1 , and p =1 2 7 Since the value of sin and cos are negative, 6 6 Thus, the respective values of and p are 7 and 1. 6 Question 3: Find the equation of the line, which cut-off intercepts on the axes whose sum and product are 1 and -6, respectively. Solution 3: Let the intercepts cut by the given lines on the axes be a and b. It is given that a + b = 1 …(1) ab = -6 …(2) On solving equations (1) and (2) , we obtain a = 3 and b = -2 or a = -2 and b = 3 It is known that the equation of the line whose intercepts on the axes are a and b is x y 1 or bx + ay – ab = 0 a b Case I: a = 3 and b = -2 In case, the equation of the line is -2x + 3y + 6 = 0, i.e., 2x – 3 y = 6. Case II: a = - 2 and b = 3 In this case, the equation of the line is 3x – 2y + 6 = 0, i.e., -3x + 2y = 6. Thus, the required equation of the lines are 2x – 3 y = 6 and -3x + 2y = 6. Question 4: x y What are the points on the y-axis whose distance from line 1is 4 units. 3 4 Solution 4: x y Let (0, b) be the point on y-axis whose distance from line 1 is 4 units. 3 4 The given line can be written as 4x + 3y – 12 = 0 …(1) On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = 4, B = 3, C = -12. 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 ______________________________________________________________________________ It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point x y is given by Ax1By 1C d 2 2 A B x y Therefore, if (0, b) is the point on the y-axis whose distance from line 1 is 4 units, then: 3 4 403b12 4 2 2 4 3 3b 12 4 5 20 3b 12 b b or b 20 3 12 20 3 12 20 3 12 3b 20 12or3b 20 12 32 8 b orb 3 3 32 Thus, the required points are 0, 3 and 8 0, 3 1, 1 Question 5: Find the perpendicular distance from the origin to the line joining the points cos ,sin and cos ,sin Solution 5: The equation of the line joining the points cos ,sin and cos ,sin cos ,sin and cos ,sin sin sin ysin xcos cos cos y is given by cos cos sin cos cos xsin sin cos sin sin y sin sin ycos cos sin( ) 0 x sin sin cos cos cos sin cos sin sin cos sin cos 0 x Ax By C 0, whereA sin sin , B cos cos , andC sin( ) It is known that the perpendicular distance (d) of a line Ax +By +C =0 from a point x y is Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor. 1, 1

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Class XI Chapter 12 - Introduction

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