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Last updated at May 29, 2023 by Teachoo

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 π¦/π₯ = (π Β± π‘πππ)/(1 β π π‘πππ) . 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 π¦/π₯ = (π Β± π‘πππ)/(1 β π π‘πππ) We know that equation of line passing through point (x1, y1) & having slope m is (y β y1) = m(x β x1) Let m1 be the slope of line OP Equation of a line OP passing through origin (0, 0) with slope m1 (y β 0) = m1(x β 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 β y1) = m(x β x1) Let m1 be the slope of line OP Equation of a line OP passing through origin (0, 0) with slope m1 (y β 0) = m1(x β 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 = |(π_2 β π_1)/(1 + π_2 π_1 )| Angle between line OP & PQ is tan π = |(π_1 β π)/(1 + π_1 π)| |(π_1 β π)/(1 + π_1 π)| = tan ΞΈ (π_1 β π)/(1 + π_1 π) = Β± tan ΞΈ So, (π_1 β π)/(1 + π_1 π) = tan ΞΈ & (π_1 β π)/(1 + π_1 π) = β tan ΞΈ Solving (π_π β π)/(π + π_π π) = tan ΞΈ (π_1 β π)/(1 + π_1 π) = tan ΞΈ m1 β m = tan ΞΈ (1 + m1m) m1 β m = tan ΞΈ + m1m tan ΞΈ m1 β m1m tan ΞΈ = m + tan ΞΈ m1 (1 β m tan ΞΈ) = m + tan ΞΈ m1 = (π + tanβ‘ΞΈ)/(1 β m tanβ‘ΞΈ ) Solving (π_π β π)/(π + π_π π) = β tan ΞΈ (π_1 β π)/(1 + π_1 π) = β tan ΞΈ m1 β m = βtan ΞΈ (1 + m1m) m1 β m = β tan ΞΈ β tan ΞΈ m1m m1 + tan ΞΈ m1m = βtan ΞΈ + m m1 (1 + m tan ΞΈ) = m - tan ΞΈ m1 = (π β tanβ‘ΞΈ)/(1 + m tanβ‘ΞΈ ) Putting value of m1 in (1) y = (π Β± tanβ‘ΞΈ)/(m β m tanβ‘ΞΈ ) π₯ π¦/π₯ = (π Β± tanβ‘ΞΈ)/(m β m tanβ‘ΞΈ ) Hence proved