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 6 3 from this point is

Question

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 $\frac{\sqrt{6}}{3}$ from this point is

Smart Achievers · Accepted Answer

(2)

$\frac{x - 1}{\cos θ} = \frac{y - 2}{\sin θ} = r = \frac{\sqrt{6}}{3} = \sqrt{\frac{2}{3}}$

The point $(r \cos θ + 1, r \sin θ + 2)$ lies on $x + y = 4$

$∴$ $r (\cos θ + \sin θ) = 1$

or $\cos θ + \sin θ = \sqrt{\frac{3}{2}}$ or $\frac{1}{\sqrt{2}} \cos θ + \frac{1}{\sqrt{2}} \sin θ = \frac{\sqrt{3}}{2}$

or $\cos (θ - 45 °) = \cos 30 °$

$∴$ $θ - 45 ° = \pm 30 °$

$∴$ $θ = 75 °, 15 ° .$

Solution

Q.
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 $\frac{\sqrt{6}}{3}$ from this point is

1 $30 °, 60 °$

2 $15 °, 75 °$

3 $∥ to x -axis$

4 $∥ to y -axis$

Want Us to Email you About Special Offers & Updates?

Solution

Q. 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 63 from this point is

1 30°, 60°

2 15°, 75°

3 ∥ to x-axis

4 ∥ to y-axis