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Lemma

TMUA practice: Straight Lines, problem 2

A TMUA-calibre straight lines problem at difficulty 4 of 5, with the full worked solution.

The question
A straight line is drawn through the point (1,2)(1, 2) making an angle θ\theta, with 0<θπ30 < \theta \le \dfrac{\pi}{3}, with the positive direction of the xx-axis, meeting the line x+y=4x + y = 4 at a point PP such that the distance of PP from (1,2)(1, 2) is 63\dfrac{\sqrt{6}}{3}. Then θ\theta equals
A π18\dfrac{\pi}{18}
B π12\dfrac{\pi}{12}
C π10\dfrac{\pi}{10}
D π3\dfrac{\pi}{3}

The correct answer is highlighted. B

Worked solution

Write the line through (1,2)(1, 2) in parametric form

(x,y)=(1+tcosθ,  2+tsinθ),(x, y) = (1 + t\cos\theta,\; 2 + t\sin\theta),

where t|t| is the distance from (1,2)(1, 2) to the point with parameter tt.

Imposing PP on x+y=4x + y = 4:

(1+tcosθ)+(2+tsinθ)=4    3+t(cosθ+sinθ)=4    t(cosθ+sinθ)=1.(1 + t\cos\theta) + (2 + t\sin\theta) = 4 \;\Longrightarrow\; 3 + t(\cos\theta + \sin\theta) = 4 \;\Longrightarrow\; t(\cos\theta + \sin\theta) = 1.

For 0<θπ30 < \theta \le \tfrac{\pi}{3} we have cosθ+sinθ>0\cos\theta + \sin\theta > 0, so t>0t > 0 and the distance equals t=63t = \dfrac{\sqrt6}{3}. Hence

cosθ+sinθ=1t=36=96=32=62.\cos\theta + \sin\theta = \frac{1}{t} = \frac{3}{\sqrt6} = \sqrt{\frac{9}{6}} = \sqrt{\frac{3}{2}} = \frac{\sqrt6}{2}.

Solving the trig equation. Using the identity cosθ+sinθ=2sin ⁣(θ+π4)\cos\theta + \sin\theta = \sqrt2\,\sin\!\left(\theta + \tfrac{\pi}{4}\right):

2sin ⁣(θ+π4)=62    sin ⁣(θ+π4)=622=32.\sqrt2\,\sin\!\left(\theta + \frac{\pi}{4}\right) = \frac{\sqrt6}{2} \;\Longrightarrow\; \sin\!\left(\theta + \frac{\pi}{4}\right) = \frac{\sqrt6}{2\sqrt2} = \frac{\sqrt3}{2}.

So θ+π4=π3\theta + \dfrac{\pi}{4} = \dfrac{\pi}{3} or 2π3\dfrac{2\pi}{3}, giving θ=π12\theta = \dfrac{\pi}{12} or θ=5π12\theta = \dfrac{5\pi}{12}.

The constraint 0<θπ3=4π120 < \theta \le \dfrac{\pi}{3} = \dfrac{4\pi}{12} admits π12\dfrac{\pi}{12} and rejects 5π12\dfrac{5\pi}{12}.

Answer: B.