# Example 11 - Chapter 9 Class 11 Straight Lines

Last updated at April 16, 2024 by Teachoo

Examples

Example 1 (a)

Example 1 (b)

Example 1 (c) Important

Example 1 (d)

Example 2 Important

Example 3 Important

Example 4 Important

Example 5

Example 6

Example 7 (i)

Example 7 (ii) Important

Example 8

Example 9 Important

Example 10

Example 11 You are here

Example 12 Important

Example 13 Important

Example 14

Example 15 Important

Example 16 Important

Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Deleted for CBSE Board 2025 Exams

Question 3 Deleted for CBSE Board 2025 Exams

Question 4 Deleted for CBSE Board 2025 Exams

Question 5 Important Deleted for CBSE Board 2025 Exams

Question 6 Important Deleted for CBSE Board 2025 Exams

Question 7 Important Deleted for CBSE Board 2025 Exams

Question 8 Deleted for CBSE Board 2025 Exams

Question 9 Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 by Teachoo

Example 11 If the lines 2x + y – 3 = 0, 5x + ky – 3 = 0 and 3x – y – 2 = 0 are concurrent, find the value of k. Three lines are concurrent if they pass through a common point i.e. point of intersection of any two lines lies on the third line It is given that lines 2x + y − 3 = 0 5x + ky − 3 = 0 3x − y − 2 = 0 are concurrent So, finding point of intersection of lines (1) & (3) Adding (1) & (3) (2x + y − 3) + (3x − y − 2) = 0 2x + 3x + y – y − 3 − 2 = 0 5x + 0 − 5 = 0 5x = 5 x = 5/5 x = 1 Putting x = 1 in (1) 2x + y − 3 = 0 2(1) + y − 3 = 0 2 + y − 3 = 0 y − 1 = 0 y = 1 Hence point of intersection of line(1) & (3) is (1, 1) Since lines (1), (2) & (3) are concurrent (1, 1) will satisfy equation of line (2) Putting x = 1 & y = 1 in (2) 5x + ky − 3 = 0 5(1) + k(1) − 3 = 0 5 + k − 3 = 0 k + 5 − 3 = 0 k + 2 = 0 k = −2 Thus, k = −2