        1. Chapter 10 Class 11 Straight Lines (Term 1)
2. Serial order wise
3. Miscellaneous

Transcript

Misc 11 Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x – 2y = 3. Let the equation line AB be x − 2y = 3 Let line CD pass through point (3, 2) & make an angle of 45° with line AB i.e. angle between CD & AB is 45° Let m1 be the slope of line AB & m2 be the slope of line CD Given angle between CD & AB is 45° i.e. θ = 45° Now, angle between two lines is given by tan θ =|(𝑚_2 − 𝑚_1)/(1 + 𝑚_2 𝑚_1 )| Here, m1 = Slope of line AB m2 = Slope of line CD & θ = angle between AB & CD = 45° Putting values tan 45° =|(𝑚_2 − 𝑚_1)/(1 + 𝑚_2 𝑚_1 )| 1 = |(𝑚_2 − 𝑚_1)/(1 + 𝑚_2 𝑚_1 )| Finding slope of line AB Given equation of line AB is x − 2y =3 −2y = 3 − x y = (3 − 𝑥)/(−2) y = 3/(−2) + (( −𝑥)/( −2)) y = (−3)/2 + 𝑥/2 y = 𝑥/2 + ((−3)/2) y = 1/2 x – 3/2 The above equation is of the form y = mx + c Where m is slope of line Slope of line AB = 1/2 i.e. m1 = 1/2 Putting m1 = 1/3 in (1) 1 = |(𝑚_2 − 1/2)/(1 + 𝑚_2 × 1/2)| 1 = |( (2𝑚_2 − 1)/2)/( (2 + 𝑚_2)/2)| 1 = |(2𝑚_2 − 1)/(2 + 𝑚_2 )| |(2𝑚_2 − 1)/(2 + 𝑚_2 )| = 1 Hence, (2𝑚_2 − 1)/(2 + 𝑚_2 ) = 1 or (2𝑚_2 − 1)/(2 + 𝑚_2 ) = −1 Solving (𝟐𝒎_𝟐 − 𝟏)/(𝟐 + 𝒎_𝟐 ) = 1 (2𝑚_2 − 1)/(2 + 𝑚_2 ) = 1 2m2 − 1 = 2 + m2 2m2 − m2 = 2 + 1 m2 = 3 Solving (𝟐𝒎_𝟐 − 𝟏)/(𝟐 + 𝒎_𝟐 ) = − 1 (2𝑚_2 − 1)/(2 + 𝑚_2 ) = − 1 2m2 − 1 = −1 (2 + m2) 2m2 − 1 = −2 − m2 2m2 + m2 = −2 + 1 3m2 = − 1 m2 = (−1)/3 So, m2 = 3 or ( − 1)/3 ∴ Slope of line CD is 3 or (−𝟏)/𝟑 Now we need to calculate equation of line CD Equation of line passes through (x1, y1) & having slope m is (y – y1) = m(x – x1) Equation of line CD passing through (3, 2) & having slope m2 is (y – 2) = m2(x – 3) When m2 = 3 (y – 2) = 3(x – 3) y – 2 = 3x − 9 y − 3x = − 9 + 2 −3x + y = −7 −(3x − y) = −7 3x − y = 7 When m2 = (−𝟏)/𝟑 (y – 2) = − 1/3 (x – 3) 3(y – 2) = −1(x – 3) 3y – 6 = −x + 3 3y + x = 3 + 6 3y + x = 9 x + 3y = 9 Hence required equation of line is 3x − y = 7 or x + 3y = 9

Miscellaneous 