Find the angles between the lines √3x + y = 1 and x + √3y = 1

Solution:

The given lines are √3x + y = 1 and x + √3y = 1.

y = - √3x + 1 ....(1)

y = (- 1/√3) x + 1/√3 ....(2)

The slope of line (1) is m\(_1\) = - 3, while the slope of the line (2) is m\(_2\) = - 1/√3

The actual angle i.e., θ between the two lines is given by

tan θ = | [m\(_1\) - m\(_2\)] / (1 + m\(_1\) m\(_2\))|

tan θ = | [(- 3) - (- 1/√3)] / (1 + (- 3) (- 1/√3)|

= |(- √3 + 1/√3) / (1 + (- √3(- 1/√3))|

= | [(- 3 + 1)/√3] / (1 + 1)|

= |- 2/(2 x √3)|

tan θ = 1/√3

θ = 30°

Thus, the angle between the given lines is either 30° or 180°- 30° = 150°

---

NCERT Solutions Class 11 Maths Chapter 10 Exercise 10.3 Question 9

Find the angles between the lines √3x + y = 1 and x + √3y = 1

Summary:

The angles between the lines √3x + y = 1 and x + √3y = 1 is either 30° or 180°- 30° = 150°