# Ex 10.3, 12 - Chapter 10 Class 11 Straight Lines (Term 1)

Last updated at Feb. 3, 2020 by Teachoo

Ex 10.3

Ex 10.3, 1 (i)

Ex 10.3, 1 (ii) Important

Ex 10.3, 1 (iii)

Ex 10.3, 2 (i)

Ex 10.3, 2 (ii)

Ex 10.3, 2 (iii) Important

Ex 10.3, 3 (i) Deleted for CBSE Board 2023 Exams

Ex 10.3, 3 (ii) Deleted for CBSE Board 2023 Exams

Ex 10.3, 3 (iii) Important Deleted for CBSE Board 2023 Exams

Ex 10.3, 4

Ex 10.3, 5 Important

Ex 10.3, 6 (i) Important

Ex 10.3, 6 (ii)

Ex 10.3, 7

Ex 10.3, 8 Important

Ex 10.3, 9 Important

Ex 10.3, 10

Ex 10.3, 11

Ex 10.3, 12 Important You are here

Ex 10.3, 13

Ex 10.3, 14 Important

Ex 10.3, 15

Ex 10.3, 16 Important

Ex 10.3, 17 Important

Ex 10.3, 18 Important

Last updated at Feb. 3, 2020 by Teachoo

Ex 10.3, 12 Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line. We know that Angle between 2 lines be tan θ =|(𝑚_2 − 𝑚_1)/(1 + 𝑚_2 𝑚_1 )| Here m1 = Slope of one line = 2 θ = 60° (given) We need to find m2 Putting the values tan 60° = |(𝑚_2 − 2)/(1 + 2 × 𝑚_2 )| √3 = |(𝑚_2 − 2)/(1 + 2𝑚_2 )| |(𝑚_2 − 2)/(1 + 2𝑚_2 )|= √3 (𝑚_2 − 2)/(1 + 2𝑚_2 ) = ± √3 So, (𝑚_2 − 2)/(1 + 2𝑚_2 ) = √3 and (𝑚_2 − 2)/(1 + 2𝑚_2 ) = – √3 Taking (𝒎_𝟐 − 𝟐)/(𝟏 + 𝟐𝒎_𝟐 ) = √𝟑 m2 − 2 = √3(1 + 2m2) m2 − 2 = √3 + 2√3m2 m2 – 2√3m2 = √3 + 2 m2 (1 – 2√3) = 2 + √3 m2 = (2 + √3)/(1 − 2√3) Taking (𝒎_𝟐 − 𝟐)/(𝟏 + 𝟐𝒎_𝟐 ) = − √𝟑 m2 − 2 = − √3(1 + 2m2) m2 − 2 = − √3 − 2√3m2 m2 + 2√3m2 = −√3 + 2 m2 (1 + 2√3) = 2 − √3 m2 = (2 − √3)/(1 + 2√3) We know that equation of a line passing through (x1, y1) & having slope m is (y − y1) = m(x − x1) Equation of a line passing through (2, 3) & having slope (𝟐 + √𝟑)/(𝟏 − 𝟐√𝟑) is (y − 3) = ( (2 + √3))/(1 − 2√3) (x − 2) (1 − 2√3)(y − 3) = (2 + √3)(x − 2) 1(y − 3) − 2√3(y − 3) = 2(x – 2) + √3(x − 2) y – 3 − 2√3y + 6√3 = 2x – 4 + √3x − 2√3 y − 2√3y − 4x − √3x = – 6√3 – 2√3 – 4 + 3 y (1 − 2√3) − x(√3 + 2) = – 1 – 8√3 1 + 8√3 = x(√3 + 2) + y (2√3 – 1) (√𝟑 + 2)x + (2√𝟑 – 1)y = 1 + 8√𝟑 Equation of a line passing through (2, 3) & having slope (𝟐 − √𝟑)/(𝟏 + 𝟐√𝟑) is (y − 3) = (2 − √3)/(2√3 + 1)(x − 2) (2√3 + 1) (y − 3) = (2 − √3) (x − 2) 2√3 (y − 3) + 1 (y − 3) = 2(x − 2) − √3(x − 2) 2√3 (y − 3) + 1 (y − 3) = 2(x − 2) − √3(x − 2) 2√3y − 6√3 + y − 3 = 2x − 4 − √3x + 2√3 2√3y + y − 6√3 − 3 = 2x − √3x − 4 + 2√3 (2√3 + 1)y − 6√3 − 3 = (2 − √3)x − 4 + 2√3 (2√3 + 1)y = − (√3 − 2)x − 4 + 2√3 + 6√3 + 3 (2√3 + 1)y + (√3−2)x = 2√3 + 6√3 − 4 + 3 (√𝟑−𝟐)x + (2√𝟑 + 1)y = 8√𝟑 − 1 Hence the equation of lines is (√3 + 2)x + (2√3 – 1)y = 1 + 8√3 or (√3 − 2)x + (2√3 + 1)y = 8√3 − 1