Question

Given a $△ABC$ with unequal sides. P is the set of all points which is equidistant from B & C and Q is the set of all point which is equidistant from sides AB and AC. Then $n(P\cap Q)$ equals:

• A. $1$
• B. $2$
• C. $3$
• D. Infinite

Answer 1

The correct option is A $1$

Laws of a point equidistant from B and C is the perpendicular bisector of BC.

Locus of a point equidistant from two intersecting lines is the angular bisector between the two lines.

$\therefore$ Angular bisector of angle between AB & AC will coincide with perpendicular bisector of BC at one point.

$\therefore n(P\cap Q)=1$

Note: If the triangle would have been isosceles or equilateral, both angular bisector line and perpendicular bisector would have overlapped,hence,$n(P\cap Q)=1$ would have been infinite.

But with unequal sides,A)1