# Misc 12

Last updated at March 9, 2017 by Teachoo

Last updated at March 9, 2017 by Teachoo

Transcript

Misc 12 Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes. Given lines are 4x + 7y − 3 = 0 2x – 3y + 1 = 0 We need to calculate Equation of line that passes through point of intersection of lines (1) & (2) & make equal intercepts on the axes Calculating point of intersection of lines (1) & (2) From (2) 2x − 3y + 1 = 0 2x = 3y − 1 x = (3𝑦 − 1)/2 Putting value of x in (1) 4x + 7y − 3 = 0 4((3𝑦 − 1)/2) + 7y − 3 = 0 2(3y − 1) + 7y − 3 = 0 6y − 2 + 7y − 3 = 0 13y − 5 = 0 13y = 5 y = 5/13 Putting value of y = 5/13 in (1) 2x − 3y + 1 = 0 2x − 3(5/13) + 1 = 0 2x − 15/13 + 1 = 0 2x = 15/13 – 1 2x = (15 − 13)/13 2x = 2/13 x = 2/(13 × 2) x = 1/13 Thus, point of intersection is (1/13, 5/13) Given that equation of line makes equal intercepts on the axes Let equation of line be 𝑥/𝑎 + 𝑦/𝑏 = 1 Where a is x − intercept & b is y − intercept Here both intercepts are same, So, b = a So equation of line becomes 𝑥/𝑎 + 𝑦/𝑎 = 1 x + y = a Also the line passes through (1/13, 5/13) Putting x = 1/13, y = 5/13 in our equation x + y = a 1/13 + 5/13 = a 6/13 = a a = 6/13 Thus, the equation of required lines becomes x + y = a x + y = 6/13 13(x + y) = 6 13x + 13y = 6 13x + 13y − 6 = 0

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .