

Get live Maths 1-on-1 Classs - Class 6 to 12
Miscellaneous
Misc 2 Important Deleted for CBSE Board 2023 Exams
Misc 3 Important
Misc 4 You are here
Misc 5
Misc 6 Important
Misc 7
Misc 8 Important
Misc 9 Important
Misc 10
Misc 11 Important
Misc 12 Important
Misc 13
Misc 14
Misc 15 Important
Misc 16 Important
Misc 17
Misc 18 Important
Misc 19 Important
Misc 20 Important
Misc 21 Important
Misc 22 Important
Misc 23
Misc 24 Important
Last updated at March 16, 2023 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)