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.

Answer 1

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

$\therefore \frac{x+1}{cos\theta }=\frac{y-2}{sin\theta }=3$

$\therefore x+1=3cos\theta$

$\Rightarrow x=3cos\theta -1$

and $y-2=3sin\theta$

$\Rightarrow y=3sin\theta +2$

Since this point lies on the line $x+y=4$

$\therefore 3cos\theta -1+3sin\theta +2=4$

$\therefore 3cos\theta +sin\theta =1$

$\Rightarrow cos\theta +sin\theta =1$

Squaring both sides, we have

$co{s}^2\theta +si{n}^2\theta +2sin\theta cos\theta =1$

$\Rightarrow 1+sin2\theta =1\Rightarrow sin2\theta =0$

$\Rightarrow 2\theta =0\Rightarrow \theta =0$

which shows that required line is parallel to x-axis or parallel to y-axis.