Mathematics [3 marks sum]

(I) Draw PQ, the perpendicular bisector of chord AC. PQ is the required locus, which is the diameter of the circle.
Reason: We know each point on the perpendicular bisector of AB is equidistant from A and B. Also the perpendicular bisector of a chord, passes through the centre of the circle and any chord passing through the centre of the circle is its diameter.

∴ PQ is the diameter of the circle.
(ii) Chords AB and AC intersects at M and N is a moving point such that LM = LN, where LM ⊥ AB and LN ⊥ AC
In right ΔALN and ΔALB
∠ANL = ∠ABL ...(90° each)
AL = AL ...(Common)
NL = BL ...[Given]
∴ ΔALN = ΔALB ...[R.H.S.]
Hence ∠MAL = ∠BAL ...c.p.c.t.
Thus, L lies on the bisector of ∠BAC.
Hence proved.

Given: A quadrilateral ABCD in which bisectors of ∠B and ∠C meet in P. PM ⊥ ABand PN ⊥ CD.
To prove: PM = PN
Construction: Draw PL ⊥ BC

Proof: Since, P lies on the bisector of ∠B
∴ P is equidistant from BC and BA
⇒ PL = PM ...(i)
Also, P lies on the bisector of ∠C ...[Given]
∴ P is equidistant from CB and CD
⇒ PL = PN ...(ii)
From (i) and (ii), we have
PL = PM
and PL = PN
⇒ PM = PN.
Hence proved.

(i) The locus of points equidistant from three given points A, B & C is the straight line PQ, which bisects AB at right angles.

(ii) Similarly, the locus of points equidistant from B and C is the straight line RS which bisects BC at right angles.
Hence, the point common to PQ and RS must satisfy both conditions; that is to say, X the point of intersection of PQ and RS will be equidistant from A, B and C.

Draw an angle bisectcr PQ and RS of angles formed by the lines m and n. From centre draw a circle with radius 2 cm, whidi intersect the angle bisectors at a, b, c and d respectively. Hence, a, b, c and d are the required four points.

Steps of construction :
(i) Draw AC = 4·9 cm, draw AB = 5·5 cm and AC = 4·9 cm.
(ii) Draw bisector l ⊥ AC.
(iii) Draw AO ⊥ AB.
(iv) Intersection of AO and L is centre of circle.

Let P and Q be the centres of two circles S and S', each passing through two given points A and B. Then,
PA = PB ...[Radii of the same circle]
⇒ P lies on the perpendicular bisector of AB ...(i)
Again, QA = QB ...[Radii of the same circle]
⇒ Q lies on the perpendicular bisector of AB ...(ii)

From (i) and (ii), it follows that P and Q both lies on the perpendicular bisector of AB.
Hence, the locus of the centres of all the circles passing through A and B is the perpendicular bisector of AB.

Let A, B and C be three non-collinear points. Join AB and BC. Let P be a moving point. Since, P is equidistant from A and B, it follows that P lies on the perpendicular bisector of AB.

Again P is equidistant from B and C. So, P lies on the perpendicular bisector of BC.
Thus, P is the point of intersection of the perpendicular bisector of AB and BC. So, P coincides three given non-collinear points. Hence, the required locus is the centre of the circle passing through three given non-collinear points.

Let A, B, C be three distinct points on a line1. Any point equidistant from A and B lies on the perpendicular bisector of Ab. So, points equidistant from A and B lies on line m.
Similarly, points equidistant from B and C lies on line n which is the perpendicular bisector of BC.
Thus, any point equidistant from A, B and C must be common to both the lines m and n.

But m ⊥ AB and n ⊥ BC.
∴ m ⊥ AC and n ⊥ AC
⇒ m || n
So, no points is common to both m and n.
Hence, the required locus is the null set Φ.

Since, the diagonals AC and BD of quadrilateral ABCD bisect each other at right angles.
∴ AC is the ⊥ bisector of line segment BD
⇒ A and C both are equidistant from B and D
⇒ AB = AD and CB = CD ...(i)

Also, BD is the ⊥ bisector of line segment AC
⇒ B and D both are equidistant from A and C
⇒ AB = BC and AD = DC ...(ii)
From (i) and (ii), we get
AB = BC = CD = AD
Thus, ABCD is a quadrilateral whose diagonals bisect each other at right angles and all four sides are equal.
Hence, ABCD is a rhombus.
Hence proved.

(i) Steps of construction:
(1) Draw BC = 6 cm.
(2) Draw ∠ABC = 120°.
(3) Cut BA = 3.5 cm.
(4) Join A to C.
(5) Draw ⊥ bisector MN of BC.
(6) Draw a circle O as centre and OC, OB radius.
(7) Draw angle bisector of ∠ABC which intersect circle at P.

(ii) ∠BCP = 30°.

Given: ΔPBC and ΔQBC are two isosceles triangles on the same base BC.
To prove: Line PQ is the perpendicular bisector of BC.
Proof: In ΔPBC, PB = PC
Since, the locus of a point equidistant from B and C is the perpendicular bisector of 1 of the line segment BC

∴ P lies on 1
Similarly Q lies on 1
Therefore, PQ is the perpendicular bisector of BC.
Hence proved.