# Misc 12 - Chapter 9 Class 11 Straight Lines

Last updated at April 16, 2024 by Teachoo

Miscellaneous

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Question 1 Important Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 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