Class XI Chapter 10 – Straight Lines Maths ______________________________________________________________________________ Question 2: Reduce the following equations into intercept form and find their intercepts on the axes. (i)3x + 2y – 12 = 0 (ii) 4x – 3y = 6 (iii) 3y + 2 =0. Solution 2: (i) The given equation is 3x – 2y – 12 = 0 It can be written as 3x2y12 3x 2y 1 12 12 x y i.e., 1 ... 1 4 6 x y This equation is of the form 1, where a = 4 and b = 6. a b Therefore, equation (1) is in the intercept form, where the intercepts on the x and y axes are 4 and 6 respectively. (ii) The given equation is 4x – 3y = 6. It can be written as 4x 3y 1 6 6 2x y 1 3 2 x y i.e., 1 ... 2 3 2 2 Therefore, equation (2) is in the intercept form, where the intercepts on x and y axes are 3 2 and -2 respectively. (iii) The given equation is 3y + 2 =0. It can be written as 3y = - 2 y i.e., = 1 ...(3) 2 3 x y Therefore, equation is in the 1, where a = 0 and b = a b 2 . 3 Therefore, equation (3) is in the intercept form, where the intercept on the y-axis is has no intercept on the x-axis. 2 and it 3 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 ______________________________________________________________________________ Question 3: Reduce the following equations into normal form. Find their perpendicular distance from the origin and angle between perpendicular and the positive x-axis. (i)x- 3y 8= 0 (ii) y – 2 = 0 (iii) x – y = 4 Solution 3: (i) The given equation is x- 3y 8= 0 It can be written as: x 3y 8 x 3y 8 2 On dividing both sides by 2 x 3 8 y 2 2 2 1 3 x y 4 2 2 xcos120 ysin120 4 ... 1 1 3 4 2, we obtain Equation (1) is in the normal form. On comparing equation (1) with the normal form of equation of the line xcos ysin p , we obtain = 120 and p =4. Thus, the perpendicular distance of the line from the origin is 4, while the angle between the perpendicular and the positive x-axis is 120 . (ii) The given equation is y – 2 = 0. It can be reduced as 0.x + 1.y = 2 On dividing both sides by 2 2 0 1 = 1, we obtain 0.x + 1.y = 2 xcos90 ysin90 2 ... 2 Equation (2) is in the normal form . On comparing equation (2) with the normal form of equation of line xcos ysin p , we obtain = 90 and p = 2. Thus, the perpendicular distance of the line from the origin is 2, while the angle between the perpendicular and the positive x-axis is 90 . (iii) The given equation is x – y = 4. It can be reduced as 1.x + (-1) y = 4 2 On dividing both sides by 2 1 1 = 2 , we obtain 1 1 4 x y 2 2 2 xcos 2 ysin 2 2 2 4 4 xcos315 ysin315 2 2 ...(3) Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

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