JEE Mathematics Concurrent Lines in a Triangle MCQs Set 01

Answer: (b) the perpendicular bisector of \(AB\)

Explanation:
1. If a circle passes through \(A\) and \(B\), its centre \(C\) satisfies \(CA=CB\) (both are radii).
2. The set of points equidistant from \(A\) and \(B\) is the perpendicular bisector of \(AB\).
3. Every point of that bisector is the centre of such a circle.

Teacher's Note:
1. Locus of points equidistant from two fixed points is their perpendicular bisector.
2. The centres lie on a line, not on a curve.

Answer: (c) rhombus

Explanation:
1. Diagonals that bisect each other make \(ABCD\) a parallelogram.
2. Because the diagonals are also perpendicular, \(A\) and \(C\) lie on the perpendicular bisector of \(BD\), so \(AB=AD\) and \(CB=CD\).
3. A parallelogram with equal adjacent sides is a rhombus.

Teacher's Note:
1. Bisecting diagonals give a parallelogram; perpendicular diagonals then give a rhombus.
2. Equal diagonals would instead give a rectangle.

Answer: (c) the circumcentre of triangle \(ABC\) (a single point)

Explanation:
1. Points equidistant from \(A\) and \(B\) lie on the perpendicular bisector of \(AB\); similarly for \(BC\) and \(CA\).
2. The three bisectors meet in exactly one point, so only that point is equidistant from all three.
3. That point is the centre of the circle through \(A\), \(B\), \(C\), the circumcentre.

Teacher's Note:
1. Three non-collinear points determine a unique circle through them.
2. If the points were collinear, no point would be equidistant from all three.

Answer: (a) \(PX+XR\)

Explanation:
1. \(X\) lies on the perpendicular bisector of \(PQ\), so \(XP=XQ\).
2. \(X\) lies on segment \(QR\), so \(QR=QX+XR\).
3. Replacing \(QX\) by \(PX\) gives \(QR=PX+XR\).

Teacher's Note:
1. Every point on a perpendicular bisector is equidistant from the two end points.
2. This trick converts a distance to \(Q\) into a distance to \(P\).

Answer: (a) \(XL=XM\)

Explanation:
1. \(X\) lies on the bisector of the angle between lines \(AB\) and \(AC\).
2. Any point on an angle bisector is equidistant from the two arms.
3. So \(XL=XM\).

Teacher's Note:
1. Points on an angle bisector are equidistant from both arms of the angle.
2. Congruence of triangles \(AXL\) and \(AXM\) (AAS) also proves it.

Answer: (d) the circumcentre

Explanation:
1. Let the bisectors of \(AB\) and \(AC\) meet at \(O\). Then \(OA=OB\) and \(OA=OC\), so \(OB=OC\).
2. Hence \(O\) also lies on the perpendicular bisector of \(BC\), so all three bisectors are concurrent at \(O\).
3. \(O\) is equidistant from the three vertices, so it is the circumcentre.

Teacher's Note:
1. The circumcentre is the centre of the circle through the three vertices.
2. Its distance from each vertex is the circumradius.

Answer: (c) the perpendicular bisectors of the sides of \(PQR\)

Explanation:
1. \(ABCQ\) and \(BCAR\) are parallelograms, so \(AQ=BC=AR\); thus \(A\) is the mid-point of \(RQ\).
2. The altitude from \(A\) is perpendicular to \(BC\) and hence to \(RQ\), so it is the perpendicular bisector of \(RQ\).
3. The three altitudes of \(ABC\) are therefore the perpendicular bisectors of \(PQR\) and meet at one point (the orthocentre of \(ABC\)).

Teacher's Note:
1. This is the standard proof that the three altitudes of a triangle are concurrent.
2. The orthocentre of \(ABC\) is the circumcentre of the larger triangle.

Answer: (b) isosceles

Explanation:
1. Let medians \(BE=CF\) meet at the centroid \(G\), which divides each median in the ratio \(2:1\).
2. So \(BG=CG\) and \(GE=GF\).
3. Triangles \(BGF\) and \(CGE\) are congruent (SAS), so \(BF=CE\), giving \(AB=AC\).

Teacher's Note:
1. Equal medians correspond to equal sides, those from which they are drawn to the other vertices.
2. Equilateral is a special case, but the conclusion that holds in general is isosceles.

Answer: (d) The circumcentre, the orthocentre, the centroid and the incentre

Explanation:
1. \(AB=AC\) and \(BD=DC\) make \(AD\) the perpendicular bisector of \(BC\).
2. Since \(D\) is the mid-point, \(AD\) is a median; it is perpendicular to \(BC\), so it is the altitude; and triangles \(ABD\) and \(ACD\) are congruent, so it bisects angle \(A\).
3. The line \(AD\) is median, altitude, angle bisector and perpendicular bisector, so all four centres lie on it.

Teacher's Note:
1. In an isosceles triangle the four special lines from the apex coincide.
2. In an equilateral triangle all four centres coincide at one point.

Answer: (d) orthocentre

Explanation:
1. Let the altitudes of \(ABC\) be \(AD\), \(BE\), \(CF\), meeting at \(P\). In triangle \(PBC\), \(AD\perp BC\) and \(P\) lies on \(AD\), so the line \(PD\) through \(A\) is the altitude from \(P\).
2. Side \(PC\) lies along \(CF\) and \(AB\perp CF\), so \(BA\) is the altitude from \(B\); likewise side \(PB\) lies along \(BE\) and \(AC\perp BE\), so \(CA\) is the altitude from \(C\).
3. All three altitudes of \(PBC\) pass through \(A\), so \(A\) is its orthocentre.

Teacher's Note:
1. The four points \(A,B,C,P\) form an orthocentric system: each is the orthocentre of the triangle formed by the other three.
2. The orthocentre of an acute triangle lies inside, but \(A\) lies outside triangle \(PBC\).