Ex 10.3, 18 - If p is length of perpendicular from origin - Slope-Intercept form

  1. Class 11
  2. Important Question for exams Class 11
Ask Download

Transcript

Ex10.3, 18 If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that 1/๐‘2 = 1/๐‘Ž2 + 1/๐‘2 . Equation of line whose intercept on the axes are a & b is ๐‘ฅ/๐‘Ž + ๐‘ฆ/๐‘ = 1 The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by d = |๐ด๐‘ฅ_1 + ๐ต๐‘ฆ_1 + ๐ถ|/โˆš(๐ด^2 + ๐ต^2 ) Now, ๐‘ฅ/๐‘Ž + ๐‘ฆ/๐‘ = 1 (1/๐‘Ž)x + (1/๐‘)y โ€“ 1 = 0 Comparing with Ax + By + C = 0 Hence, A = 1/๐‘Ž, B = 1/๐‘ & C = โ€“1 Also, Distance from origin (0, 0) to the line ๐‘ฅ/๐‘Ž + ๐‘ฆ/๐‘ = 1 is p So, distance d = p & x1 = 0, y1 = 0 Putting values d = |๐ด๐‘ฅ_1 + ๐ต๐‘ฆ_1 + ๐ถ|/โˆš(๐ด^2 + ๐ต^2 ) p = |(0)(1/๐‘Ž) + (0)(1/๐‘) โˆ’ 1|/โˆš((1/๐‘Ž)^2 + (1/๐‘)^2 ) p = |0 + 0 โˆ’ 1|/โˆš(1/๐‘Ž2 + 1/๐‘2) p = (| โˆ’ 1|)/โˆš(1/๐‘Ž2 + 1/๐‘2) p = 1/โˆš(1/๐‘Ž2 + 1/๐‘2) 1/๐‘ = โˆš(1/๐‘Ž2 + 1/๐‘2) Squaring both sides (1/๐‘)^2 = (โˆš(1/๐‘Ž2 + 1/๐‘2))^2 1/๐‘2 = 1/๐‘Ž2 + 1/๐‘2 Hence proved

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
Jail