NCERT Solutions
Class 11 Maths
Straight Lines

Ex10.2 Q.4
Find the equation of the line which passes though (2, 2√3) and is inclined with the x-axis at an angle of 75°.
The slope of the line that inclines with the x-axis at an angle of 75° is
m = tan 75°
=> m = tan (45° + 30°)
=> m = (tan 45° + tan 30°) ÷ (1 - tan 45° × tan 30°)
=> m = (1 + (1 ÷ √3)) ÷ (1 - 1 × (1 ÷ √3))
=> m = (1 + (1 ÷ √3)) ÷ (1 – (1 ÷ √3))
=> m = (√3 + 1) ÷ (√3 - 1)
We know that the equation of the line passing through point (x0, y0), whose slope is m, is
y – y0 = m (x – x0)
Thus, if a line passes though (2, 2√3) and inclines with the x-axis at an angle of 75°, then the equation of the line is given as
(y - 2√3) = {(√3 + 1) ÷ (√3 - 1)} (x - 2)
=> (y - 2√3) (√3 - 1) = (√3 + 1) (x - 2)
=> y (√3 - 1) - 2√3(√3 - 1) = x (√3 + 1) - 2(√3 + 1)
=> y (√3 - 1) - 6 + 2√3 = x (√3 + 1) - 2√3 – 2
=> x (√3 + 1) - 2√3 – 2 - y (√3 - 1) + 6 - 2√3 = 0
=> x (√3 + 1) - y (√3 - 1) + 4 - 4√3 = 0
=> x (√3 + 1) - y (√3 - 1) + 4(1 - √3) = 0