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Last updated at May 29, 2018 by Teachoo

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Misc 4 What are the points on the y-axis whose distance from the line ๐ฅ/3 + ๐ฆ/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 ๐ฅ/3 + ๐ฆ/4 = 1 is 4 units Given line is ๐ฅ/3 + ๐ฆ/4 = 1 (4๐ฅ + 3๐ฆ)/12 = 1 4x + 3y = 12 4x + 3y โ 12 = 0 The above equation is of the form Ax + By + C = 0 Here A = 4, B = 3, and C = โ12 We know that Distance of a point (x1, y1) from a line Ax + By + C = 0 is d = |ใ๐ด๐ฅใ_1 + ใ๐ต๐ฆใ_1 + ๐ถ|/โ(๐ด^2 + ๐ต^2 ) Given Distance of a point (0, k) from line 4x + 3y โ 12 = 0 is 4 Putting values x1 = 0 , y1 = k , d = 4 & A = 4 , B = 3 , C = โ 12 So, 4 = |4(0) + 3๐ + ( โ 12)|/โ((4)^2 + (3)^2 ) 4 = |0 + 3๐ โ 12|/โ(16 + 9) 4 = |3๐ โ 12|/โ25 4 = |3๐ โ 12|/5 4 ร 5 = |3๐โ12| 20 = |3๐โ12| |3๐โ12| = 20 3k โ 12 = ยฑ 20 Hence point on y-axis is (0, 32/3) and (0, ( โ 8)/3)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.