$A$ is a point on either of the two lines $y+\sqrt3|x|=2$ at a distance of $\frac{4}{\sqrt3}$ units from their point of intersection.

Question

$A$ is a point on either of the two lines $y+\sqrt3|x|=2$ at a distance of $\frac{4}{\sqrt3}$ units from their point of intersection.Find the coordinates of the foot of perpendicular from $A$ on the bisector of the angle between them?

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The two lines are $y+\sqrt3x=2$ and $y-\sqrt3x=2$.Their point of intersection is $(0,2)$ and the angle bisectors,i found are $x=0$(acute angle bisector) and $y=2$(obtuse angle bisector) using the angle bisector formula.

But i could not solve it further.Please help me.

Answer 1

So, any point on the $y+\sqrt3|x|=2$ will be $\left(\pm\dfrac{2-a}{\sqrt3},a\right)$

The distance of $\left(\pm\dfrac{2-a}{\sqrt3},a\right)$ from $(0,2)$

$$=\sqrt{\dfrac{(2-a)^2}3+(a-2)^2}=\dfrac2{\sqrt3}|a-2|$$ which needs to be $\dfrac4{\sqrt3}$

Answer 2

For foot of perpendicular we have a shortcut formula its (x-x1)a =(y-y1)b=k(ax1+by1+c)/a+b+c and for foot of perpendicular substitute k=1 and note here a is 1 and b is + or - (√3) separately solve for x and y and u have the point. Or use the distance formula