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Misc 13 - Equation of line passing through origin, making angle

Misc 13 - Chapter 10 Class 11 Straight Lines - Part 2
Misc 13 - Chapter 10 Class 11 Straight Lines - Part 3
Misc 13 - Chapter 10 Class 11 Straight Lines - Part 4
Misc 13 - Chapter 10 Class 11 Straight Lines - Part 5

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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 12 years. He provides courses for Maths and Science at Teachoo.