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Straight Lines — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsStraight Lines5 Marks
Question
Find the direction in which a straight line must be drawn through the point $(-1, 2)$ so that its point of intersection with the line $x + y = 4$ may be at a distance of $3$ units from this point.
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Answer

Let the required line makes an angle \theta with the positive direction of $x-$axis. Then equation of line is
$\frac{{x - ( - 1)}}{{\cos \theta }} = \frac{{y - 2}}{{\sin \theta }} = r \Rightarrow \frac{{x + 1}}{{\cos \theta }} = \frac{{y - 2}}{{\sin \theta }} = r$
It is given that $r = 3$
$\therefore \frac{{x + 1}}{{\cos \theta }} = \frac{{y - 2}}{{\sin \theta }} = 3$
$\therefore x + 1 = 3 \cos \theta \Rightarrow x = 3 \cos \theta - 1$
and $y - 2 = 3 \sin \theta \Rightarrow y = 3 \sin \theta + 2$
Since this point on the line $x + y = 4$
$\therefore 3 \cos \theta - 1 + 3 \sin \theta + 2 = 4$
$\therefore 3 \cos \theta + 3 \sin \theta = 3 \Rightarrow \cos \theta + \sin \theta = 1$
Squaring both sides, we have
$\cos^2 \theta + \sin^2 \theta + 2 \sin \theta \cos \theta = 1$
$\Rightarrow 1 + \sin 2 \theta = 1 \Rightarrow \sin 2\theta = 0 \Rightarrow 2\theta = 0 \Rightarrow \theta = 0$
Which shows that required line is parallel to $x-$axis .

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