Ex 9.3, 10 - Chapter 9 Class 11 Straight Lines
Last updated at April 16, 2024 by Teachoo
Ex 9.3
Ex 9.3, 1 (ii) Important
Ex 9.3, 1 (iii)
Ex 9.3, 2 (i)
Ex 9.3, 2 (ii)
Ex 9.3, 2 (iii) Important
Ex 9.3, 3
Ex 9.3, 4 Important
Ex 9.3, 5 (i) Important
Ex 9.3, 5 (ii)
Ex 9.3, 6
Ex 9.3, 7 Important
Ex 9.3, 8 Important
Ex 9.3, 9
Ex 9.3, 10 You are here
Ex 9.3, 11 Important
Ex 9.3, 12
Ex 9.3, 13 Important
Ex 9.3, 14
Ex 9.3, 15 Important
Ex 9.3, 16 Important
Ex 9.3, 17 Important
Question 1 (i) Deleted for CBSE Board 2025 Exams
Question 1 (ii) Deleted for CBSE Board 2025 Exams
Question 1 (iii) Important Deleted for CBSE Board 2025 Exams
Last updated at April 16, 2024 by Teachoo
Ex 9.3, 11 Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is A (x – x1) + B (y – y1) = 0. The line passing through (x1, y1) and parallel to the line Ax + By + C = 0 has the same slope as the line Ax + By + C = 0 Finding slope of Ax + By + C = 0 By = − Ax − C y = ( − 𝐴𝑥 − 𝐶)/𝐵 y = (( − 𝐴)/𝐵) x + (( −𝐶)/𝐵) The above equation is of the form y = mx + c Where m = slope of line Slope of line (Ax + By + C = 0) = ( − 𝐴)/𝐵 Since both lines are parallel Slope of the required line also same Hence, Slope of required line = − 𝐴/𝐵 Now, finding equation of line We know that equation of a line passing through the point (x1, y1) & having slope m is (y – y1) = m (x – x1) So, equation of line passing through point (x1, y1) & having slope (−𝐴)/𝐵 is (y – y1) = − 𝐴/𝐵(x – x1) B (y – y1) = − A (x – x1) By − By1 = − Ax + Ax1 Ax − Ax1 + By − By1 = 0 A(x – x1) + B(y – y1) = 0 Which is the required equation Hence proved