Example 3 - Chapter 9 Class 11 Straight Lines
Last updated at April 16, 2024 by Teachoo
Examples
Example 1 (b)
Example 1 (c) Important
Example 1 (d)
Example 2 Important
Example 3 Important You are here
Example 4 Important
Example 5
Example 6
Example 7 (i)
Example 7 (ii) Important
Example 8
Example 9 Important
Example 10
Example 11
Example 12 Important
Example 13 Important
Example 14
Example 15 Important
Example 16 Important
Question 1
Question 2
Question 3
Question 4
Question 5 Important
Question 6 Important
Question 7 Important
Question 8
Question 9
Last updated at April 16, 2024 by Teachoo
Example 3 Line through the points (–2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x. Let points be A(–2, 6), B(4, 8) , C(8, 12) and D(x, 24) If two lines are perpendicular , then product of their slope is –1 So, Slope of AB × Slope of CD = –1 We know that slope of a line through the points (x1, y1) , (x2, y2)is m = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 ) Slope of line AB passing through A(– 2, 6) & B(4, 8) Here x1 = −2, y1 = 6 x2 = 4, y2 = 8 Putting avalues Slope of AB = (8 − 6)/(4 − (−2)) = 2/(4 − (−2)) = 2/6 = 1/3 Slope of line CD passing through C(8, 12) & D(x, 24) Here x1 = 8, y1 = 12 x2 = x, y2 = 24 Putting values Slope of CD = (24 − 12)/(𝑥 − 8) = 12/(𝑥 − 8) From (1) Slope of AB × Slope of CD = –1 1/3 × (12/(𝑥 − 8)) = –1 4/((𝑥 − 8)) = –1 4 = –1(x – 8) 4 = –x + 8 x = 8 – 4 x = 4 Thus, value of x is 4