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  1. Chapter 10 Class 11 Straight Lines
  2. Serial order wise

Transcript

Misc 13 Show that the equation of the line passing through the origin and making an angle ΞΈ with the line y = mx + c is 𝑦/π‘₯ = (π‘š Β± π‘‘π‘Žπ‘›πœƒ)/(1 βˆ“ π‘š π‘‘π‘Žπ‘›πœƒ) . Let OP be the line passing through origin Let PQ be the line y = mx + c Whose slope is m and makes an angle ΞΈ with line OP We need to show equation line OP is 𝑦/π‘₯ = (π‘š Β± π‘‘π‘Žπ‘›πœƒ)/(1 βˆ“ π‘š π‘‘π‘Žπ‘›πœƒ) We know that equation of line passing through point (x1, y1) & having slope m is (y – y1) = m(x – x1) Let m1 be the slope of line OP Equation of a line OP passing through origin (0, 0) with slope m1 (y – 0) = m1(x – 0) y = m1x Now we need to find slope (i.e. m1) of line OP It is given that line OP makes an angle ΞΈ with line PQ y = mx + c We know that equation of line passing through point (x1, y1) & having slope m is (y – y1) = m(x – x1) Let m1 be the slope of line OP Equation of a line OP passing through origin (0, 0) with slope m1 (y – 0) = m1(x – 0) y = m1x Now we need to find slope (i.e. m1) of line OP It is given that line OP makes an angle ΞΈ with line PQ y = mx + c Angle between two line whose slope are m1 & m2 is tan πœƒ = |(π‘š_2 βˆ’ π‘š_1)/(1 + π‘š_2 π‘š_1 )| Angle between line OP & PQ is tan πœƒ = |(π‘š_1 βˆ’ π‘š)/(1 + π‘š_1 π‘š)| |(π‘š_1 βˆ’ π‘š)/(1 + π‘š_1 π‘š)| = tan ΞΈ (π‘š_1 βˆ’ π‘š)/(1 + π‘š_1 π‘š) = Β± tan ΞΈ So, (π‘š_1 βˆ’ π‘š)/(1 + π‘š_1 π‘š) = tan ΞΈ & (π‘š_1 βˆ’ π‘š)/(1 + π‘š_1 π‘š) = – tan ΞΈ Solving (π’Ž_𝟏 βˆ’ π’Ž)/(𝟏 + π’Ž_𝟏 π’Ž) = tan ΞΈ (π‘š_1 βˆ’ π‘š)/(1 + π‘š_1 π‘š) = tan ΞΈ m1 – m = tan ΞΈ (1 + m1m) m1 – m = tan ΞΈ + m1m tan ΞΈ m1 – m1m tan ΞΈ = m + tan ΞΈ m1 (1 – m tan ΞΈ) = m + tan ΞΈ m1 = (π‘š + tan⁑θ)/(1 βˆ’ m tan⁑θ ) Solving (π’Ž_𝟏 βˆ’ π’Ž)/(𝟏 + π’Ž_𝟏 π’Ž) = – tan ΞΈ (π‘š_1 βˆ’ π‘š)/(1 + π‘š_1 π‘š) = – tan ΞΈ m1 – m = –tan ΞΈ (1 + m1m) m1 – m = – tan ΞΈ – tan ΞΈ m1m m1 + tan ΞΈ m1m = βˆ’tan ΞΈ + m m1 (1 + m tan ΞΈ) = m - tan ΞΈ m1 = (π‘š βˆ’ tan⁑θ)/(1 + m tan⁑θ ) Putting value of m1 in (1) y = (π‘š Β± tan⁑θ)/(m βˆ“ m tan⁑θ ) π‘₯ 𝑦/π‘₯ = (π‘š Β± tan⁑θ)/(m βˆ“ m tan⁑θ ) Hence proved

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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.