# Misc 3 - Chapter 9 Class 11 Straight Lines

Last updated at April 16, 2024 by Teachoo

Miscellaneous

Misc 1
Important

Misc 2 Important

Misc 3 You are here

Misc 4

Misc 5 Important

Misc 6

Misc 7 Important

Misc 8 Important

Misc 9

Misc 10 Important

Misc 11 Important

Misc 12

Misc 13

Misc 14 Important

Misc 15 Important

Misc 16

Misc 17 Important

Misc 18 Important

Misc 19 Important

Misc 20 Important

Misc 21 Important

Misc 22

Misc 23 Important

Question 1 Important Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 by Teachoo

Misc 4 What are the points on the y-axis whose distance from the line 𝑥/3 + 𝑦/4 = 1 is 4 units. Let any point on y-axis be P(0, k) Given that distance of point on y-axis from the line 𝑥/3 + 𝑦/4 = 1 is 4 units Given line is 𝑥/3 + 𝑦/4 = 1 (4𝑥 + 3𝑦)/12 = 1 4x + 3y = 12 4x + 3y − 12 = 0 The above equation is of the form Ax + By + C = 0 Here A = 4, B = 3, and C = –12 We know that Distance of a point (x1, y1) from a line Ax + By + C = 0 is d = |〖𝐴𝑥〗_1 + 〖𝐵𝑦〗_1 + 𝐶|/√(𝐴^2 + 𝐵^2 ) Given Distance of a point (0, k) from line 4x + 3y – 12 = 0 is 4 Putting values x1 = 0 , y1 = k , d = 4 & A = 4 , B = 3 , C = − 12 So, 4 = |4(0) + 3𝑘 + ( − 12)|/√((4)^2 + (3)^2 ) 4 = |0 + 3𝑘 − 12|/√(16 + 9) 4 = |3𝑘 − 12|/√25 4 = |3𝑘 − 12|/5 4 × 5 = |3𝑘−12| 20 = |3𝑘−12| |3𝑘−12| = 20 3k – 12 = ± 20 Hence point on y-axis is (0, 32/3) and (0, ( − 8)/3)