# Ex 9.3, 3 - Chapter 9 Class 11 Straight Lines

Last updated at April 16, 2024 by Teachoo

Ex 9.3

Ex 9.3, 1 (i)

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 You are here

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

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 2024 Exams

Question 1 (ii) Deleted for CBSE Board 2024 Exams

Question 1 (iii) Important Deleted for CBSE Board 2024 Exams

Last updated at April 16, 2024 by Teachoo

Ex 9.3, 3 Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2). The distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is d = |𝐴𝑥_1 + 𝐵𝑦_1 + 𝐶|/√(𝐴^2 + 𝐵^2 ) The given line is 12(x + 6) = 5(y – 2) 12x + 12 × 6 = 5y – 5 × 2 12x + 72 = 5y – 10 12x – 5y + 82 = 0 The above equation is of the form Ax + By + C = 0 where A = 12, B = –5, and C = 82 Now We have to find distance of a point (−1, 1) from a line So, x1 = –1 & y1 = 1 Finding distance d = |𝐴𝑥_1 + 𝐵𝑦_1 + 𝐶|/√(𝐴^2 + 𝐵^2 ) Putting values d = | − 12 − 5 + 82|/√((12)2 + ( − 5)2) d = | − 12 − 5 + 82|/√(144 + 25) d = 65/√169 d = 65/√(13 × 13) d = 65/13 d = 5 Thus, Required distance = 5 units