Question

P₁, P2 are points on either of the two lines ${\displaystyle -\sqrt{3}\left|x\right|}$ = 2 at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from P₁, P₂ on the bisector of the angle between the given lines.

Answer 1

Given lines are ${\displaystyle -\sqrt{3}\left|x\right|}$ = 2

⇒ ${\displaystyle y-\sqrt{3}x}$ = 2, if x ≥ 0   .....(i)

And ${\displaystyle y+\sqrt{3}x}$ = 2, if x < 0   ......(ii)

Slope of equation (i) is tan θ = ${\displaystyle \sqrt{3}}$

∴ θ = 60°

Slope of equation (ii) is tan q  ${\displaystyle -\sqrt{3}}$

∴ θ = 120°

Solving equation (i) and equation (ii) we get

${\displaystyle y-\sqrt{3}=2}$
${\displaystyle y+\sqrt{3}x=2}$
             2y = 4

⇒ y = 2

Putting the value of y is eq. (i) we get

x = 0

∴ Point of intersection of line (i) and (ii) is Q(0, 2)

∴ QO = 2

In ΔPEQ,

cos 30° = ${\displaystyle \frac{\text{PQ}}{\text{QE}}}$

${\displaystyle \frac{\sqrt{3}}{2}=\frac{\text{PQ}}{5}}$

∴ PQ = ${\displaystyle \frac{5\sqrt{3}}{2}}$

∴ OP = OQ + PQ

= ${\displaystyle 2+\frac{5\sqrt{3}}{2}}$

Hence, the coordinates of the foot of perpendicular = ${\displaystyle (0,2+\frac{5\sqrt{3}}{02})}$.