Misc 12 - Chapter 9 Class 11 Straight Lines
Last updated at April 16, 2024 by Teachoo
Angle between two lines
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 ฮธ with the line y = mx + c is ๐ฆ/๐ฅ = (๐ ยฑ ๐ก๐๐๐)/(1 โ ๐ ๐ก๐๐๐) . Let OP be the line passing through origin Let PQ be the line y = mx + c Whose slope is m and makes an angle ฮธ with line OP We need to show equation line OP is ๐ฆ/๐ฅ = (๐ ยฑ ๐ก๐๐๐)/(1 โ ๐ ๐ก๐๐๐) We know that equation of line passing through point (x1, y1) & having slope m is (y โ y1) = m(x โ x1) Let m1 be the slope of line OP Equation of a line OP passing through origin (0, 0) with slope m1 (y โ 0) = m1(x โ 0) y = m1x Now we need to find slope (i.e. m1) of line OP It is given that line OP makes an angle ฮธ with line PQ y = mx + c We know that equation of line passing through point (x1, y1) & having slope m is (y โ y1) = m(x โ x1) Let m1 be the slope of line OP Equation of a line OP passing through origin (0, 0) with slope m1 (y โ 0) = m1(x โ 0) y = m1x Now we need to find slope (i.e. m1) of line OP It is given that line OP makes an angle ฮธ with line PQ y = mx + c Angle between two line whose slope are m1 & m2 is tan ๐ = |(๐_2 โ ๐_1)/(1 + ๐_2 ๐_1 )| Angle between line OP & PQ is tan ๐ = |(๐_1 โ ๐)/(1 + ๐_1 ๐)| |(๐_1 โ ๐)/(1 + ๐_1 ๐)| = tan ฮธ (๐_1 โ ๐)/(1 + ๐_1 ๐) = ยฑ tan ฮธ So, (๐_1 โ ๐)/(1 + ๐_1 ๐) = tan ฮธ & (๐_1 โ ๐)/(1 + ๐_1 ๐) = โ tan ฮธ Solving (๐_๐ โ ๐)/(๐ + ๐_๐ ๐) = tan ฮธ (๐_1 โ ๐)/(1 + ๐_1 ๐) = tan ฮธ m1 โ m = tan ฮธ (1 + m1m) m1 โ m = tan ฮธ + m1m tan ฮธ m1 โ m1m tan ฮธ = m + tan ฮธ m1 (1 โ m tan ฮธ) = m + tan ฮธ m1 = (๐ + tanโกฮธ)/(1 โ m tanโกฮธ ) Solving (๐_๐ โ ๐)/(๐ + ๐_๐ ๐) = โ tan ฮธ (๐_1 โ ๐)/(1 + ๐_1 ๐) = โ tan ฮธ m1 โ m = โtan ฮธ (1 + m1m) m1 โ m = โ tan ฮธ โ tan ฮธ m1m m1 + tan ฮธ m1m = โtan ฮธ + m m1 (1 + m tan ฮธ) = m - tan ฮธ m1 = (๐ โ tanโกฮธ)/(1 + m tanโกฮธ ) Putting value of m1 in (1) y = (๐ ยฑ tanโกฮธ)/(m โ m tanโกฮธ ) ๐ฅ ๐ฆ/๐ฅ = (๐ ยฑ tanโกฮธ)/(m โ m tanโกฮธ ) Hence proved