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Misc 4 - What points on y-axis whose distance from x/3 + y/4 = 1 - Distance of a point from a line

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Misc 4 What are the points on the y-axis whose distance from the line π‘₯/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 π‘₯/3 + 𝑦/4 = 1 is 4 units Given line is π‘₯/3 + 𝑦/4 = 1 (4π‘₯ + 3𝑦)/12 = 1 4x + 3y = 12 4x + 3y βˆ’ 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 = |〖𝐴π‘₯γ€—_1 + 〖𝐡𝑦〗_1 + 𝐢|/√(𝐴^2 + 𝐡^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 = βˆ’ 12 So, 4 = |4(0) + 3π‘˜ + ( βˆ’ 12)|/√((4)^2 + (3)^2 ) 4 = |0 + 3π‘˜ βˆ’ 12|/√(16 + 9) 4 = |3π‘˜ βˆ’ 12|/√25 4 = |3π‘˜ βˆ’ 12|/5 4 Γ— 5 = |3π‘˜βˆ’12| 20 = |3π‘˜βˆ’12| |3π‘˜βˆ’12| = 20 3k – 12 = Β± 20 Hence point on y-axis is (0, 32/3) and (0, ( βˆ’ 8)/3)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.