Question

A line segment $UV$ of length $6\text{units}$ subtend equal angles at two points A and B lying on the same side of the line segment. A perpendicular of length $4\text{units}$ is dropped on the line segment from a fixed point whose distances from points $A,B,U\&V$ are same. Find the sum of distance of the points A and B from the fixed point.

• A. $10\text{units}$
• B. $5\text{units}$

Answer 1

The correct option is A $10\text{units}$

We know if the line segment joining two points subtends equal angles at two other points lying on the same side of the line, then all the four points lie on a circle.

Fixed point having same distance from four points $A,B,U\&V$ is centre of the circle i.e. P.

The position of the elements can be depicted as follows:



$\therefore$ A and B lies on the circle, and the required distance is equal to radius of the circle.

PR is perpendicular to the chord UV from the center.
$\therefore$ PR will bisect UV at R.
$\Rightarrow UR=\frac{UV}{2}=\frac{6}{2}=3\text{units}$

Considering the right triangle $△UPR,$
$U{P}^2=U{R}^2+P{R}^2$
$\Rightarrow UP=\sqrt{3^2+4^2}=5\text{units}=\text{radius of the circle}$

$\therefore$ The sum of distance of the points A and B form the center of the circle $=PA\left(\text{radius}\right)+PB\left(\text{radius}\right)$
$=5+5\text{units}$
$=10\text{units}$