Loci & its Contructions — Exercise 16(A)

P is a point on perpendicular bisector of side BC of △ABC, then :

1. PA = PB

2. PB = PC

3. PA = PC

4. none of these

Answer

We know that,

Locus of a point equidistant from two given points is the perpendicular bisector of the line joining the two points.

Given,

P is a point on perpendicular bisector of side BC of △ABC.

∴ PB = PC.

Hence, Option 2 is the correct option.

P is a point on bisector of angle A of △ABC. Then:

1. P is equidistant from the vertices B and C

2. PA = PC

3. PA = PB

4. P is equidistant from sides AB and AC

Answer

We know that,

The locus of a point equidistant from two intersecting lines is the bisector of the angles between the lines.

Given,

P is a point on bisector of angle A of △ABC.

∴ P is equidistant from sides AB and AC of the △ABC.

Hence, Option 4 is the correct option.

The perpendicular bisector of side AB and bisector of angle A of △ABC meet at point P. Then :

1. PA = PB

2. PA = PC

3. PB = PC

4. PB bisects ∠ABC

Answer

We know that,

Locus of a point equidistant from two given points is the perpendicular bisector of the line joining the two points.

Since,

The perpendicular bisector of side AB and bisector of angle A of △ABC meet at point P.

∴ PA = PB.

Hence, Option 1 is the correct option.

Using the information in the given diagram, state if :

1. AD = DC

2. BD = DC

3. CD bisects angle ACB

4. angle CAD is greater than angle DAB

Answer

From figure,

AD = DB = AC = BC = x (let)

In △ ADC and △ BDC,

⇒ AD = DB (Both equal to x)

⇒ AC = BC (Both equal to x)

⇒ CD = CD (Common side)

∴ △ ADC ≅ △ BDC (By S.S.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

∴ ∠ACD = ∠BCD

∴ CD bisects angle ACB.

Hence, Option 3 is the correct option.

Bisector of angle B of triangle ABC intersects side AC at point P, then point P is :

1. equidistant from vertices A and C

2. PA = PB

3. PB = PC

4. equidistant from sides AB and BC

Answer

We know that,

The locus of a point equidistant from two intersecting lines is the bisector of the angles between the lines.

Given,

Bisector of angle B of triangle ABC intersects side AC at point P.

∴ P is equidistant from sides AB and BC.

Hence, Option 4 is the correct option.

Given: PQ is perpendicular bisector of side AB of the triangle ABC.

Prove: Q is equidistant from A and B.

Answer

Join A and Q as shown in the figure below:

In ∆AQP and ∆BQP,

⇒ AP = BP [Given]

⇒ ∠QPA = ∠QPB [Each 90°]

⇒ PQ = PQ [Common]

∴ ∆AQP ≅ ∆BQP by SAS axiom.

∴ AQ = BQ [By C.P.C.T.]

Hence, proved that Q is equidistant from A and B.

Given: CP is the bisector of angle C of ∆ABC.

Prove: P is equidistant from AC and BC.

Answer

From P, draw PL ⊥ AC and PM ⊥ CB

In ∆LPC and ∆MPC,

⇒ ∠PLC = ∠PMC [Each 90°]

⇒ ∠PCL = ∠MCP [Since, CP is bisector of angle C]

⇒ PC = PC [Common]

∴ ∆LPC ≅ ∆MPC by AAS axiom.

∴ PL = PM [By C.P.C.T.]

Hence, proved that P is equidistant from AC and BC.

Given: AX bisects angle BAC and PQ is perpendicular bisector of AC which meets AX at point Y.

Prove :

(i) X is equidistant from AB and AC.

(ii) Y is equidistant from A and C.

Answer

From X, draw XL ⊥ AC and XM ⊥ AB and join YC.

(i) In ∆AXL and ∆AXM,

⇒ ∠XAL = ∠XAM [Since, AX bisects angle BAC]

⇒ AX = AX [Common]

⇒ ∠XLA = ∠XMA [Each 90°]

∴ ∆AXL ≅ ∆AXM by AAS axiom.

∴ XL = XM [By C.P.C.T.]

Hence, proved that X is equidistant from AC and AB.

(ii) In ∆YTA and ∆YTC,

⇒ AT = CT [because PQ is perpendicular bisector of AC]

⇒ ∠YTA = ∠YTC [Each 90°]

⇒ YT = YT [Common]

∴ ∆YTA ≅ ∆YTC by SAS axiom.

∴ YA = YC [By C.P.C.T.]

Hence, proved that Y is equidistant from A and C.

Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.

Answer

Steps of Construction:

1. Draw a line segment BC = 6.3 cm

2. With centre B and radius 4.2 cm, draw an arc.

3. With centre C and radius 5 cm, draw another arc which intersects the first arc at A.

4. Join AB and AC. Then, ∆ABC is the required triangle.

5. Draw LM, perpendicular bisector of BC.

6. Let LM intersect AC at D and BC at E.

7. Join DB.

In ∆DBE and ∆DCE,

⇒ BE = EC [LM is the perpendicular bisector of BC]

⇒ ∠DEB = ∠DEC [Each 90°]

⇒ DE = DE [Common]

∴ ∆DBE ≅ ∆DCE [By SAS]

∴ DB = DC [By C.P.C.T.]

Hence, proved that D is equidistant from B and C.

In each of the given figures; PA = PB and QA = QB.

(i)

(ii)

Prove in each case, that PQ (produce, if required) is perpendicular bisector of AB.

Hence, state the locus of the points equidistant from two given fixed points.

Answer

(i) Join PQ which meets AB in D.

Given, PA = PB.

∴ P is equidistant from A and B. Thus, P lies on the perpendicular bisector of AB.

Given, QA = QB.

∴ Q is equidistant from A and B. Thus, Q lies on perpendicular bisector of AB.

Thus, both P and Q lie on the perpendicular bisector of AB.

Hence, proved that PQ is the perpendicular bisector of AB.

(ii) Join PQ which meets AB in D.

Given, PA = PB.

∴ P is equidistant from A and B. Thus, P lies on the perpendicular bisector of AB.

Given, QA = QB.

∴ Q is equidistant from A and B. Thus, Q lies on perpendicular bisector of AB.

Thus, both P and Q lie on the perpendicular bisector of AB.

Hence, proved that PQ is the perpendicular bisector of AB.

Hence, locus of the points which are equidistant from two fixed points, is the perpendicular bisector of the line joining the fixed points.

Construct a triangle ABC in which angle ABC = 75°, AB = 5 cm and BC = 6.4 cm. Draw perpendicular bisector of side BC and also the bisector of angle ACB. If these bisectors intersect each other at point P; prove that P is equidistant from B and C; and also from AC and BC.

Answer

Steps of Construction:

1. Draw a line segment BC = 6.4 cm

2. At B, draw a ray BX making an angle of 75° and cut off BA = 5 cm.

3. Join AC. ∆ABC is the required triangle.

4. Draw the perpendicular bisector of BC, intersecting BC at Q.

5. Draw CX, the angle bisector of angle ACB which intersects the perpendicular bisector of BC at P.

6. Join PB and draw PL ⊥ AC.

In ∆PBQ and ∆PCQ,

⇒ PQ = PQ [Common]

⇒ ∠PQB = ∠PQC [Each = 90°]

⇒ BQ = QC [As PQ is the perpendicular bisector of BC]

∴ ∆PBQ ≅ ∆PCQ [By SAS axiom]

∴ PB = PC [By C.P.C.T.]

∴ P is equidistant from B and C.

Also,

In ∆PQC and ∆PLC,

⇒ ∠PQC = ∠PLC [Each 90°]

⇒ ∠PCQ = ∠PCL [As CX is angle bisector of ∠ACB]

⇒ PC = PC [Common]

∴ ∆PQC ≅ ∆PLC by AAS axiom.

∴ PQ = PL [By C.P.C.T.]

∴ P is equidistant from AC and BC.

Hence, proved that P is equidistant from B and C and also from AC and BC.

In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B.

Prove that P is equidistant from AB and BC.

Answer

Steps of construction:

1. Draw a parallelogram such that AB > BC.

2. Join AC the diagonal of parallelogram.

3. Draw BX, the angle bisector of ∠ABC and let it intersect AC at P.

4. From P, draw PL ⊥ AB and PM ⊥ BC.

In ∆PLB and ∆PMB,

⇒ ∠PLB = ∠PMB [Each 90°]

⇒ ∠PBL = ∠PBM [As BX is angle bisector of ABC.]

⇒ PB = PB [Common]

∴ ∆PLB ≅ ∆PMB by AAS axiom.

∴ PL = PM

Hence, proved that P is equidistant from AB and BC.

In triangle LMN, bisectors of interior angles at L and N intersect each other at point A. Prove that:

(i) point A is equidistant from all the three sides of the triangle.

(ii) AM bisects angle LMN.

Answer

Steps of construction :

1. Construct a triangle LMN.

2. Draw angle bisectors of L and N. Let the angle bisectors meet at A.

3. Join AM.

(i) Since, A lies on bisector of ∠N

∴ A is equidistant from MN and LN.

Again, as A lies on the bisector of ∠L

∴ A is equidistant from LN and LM.

Hence, proved that A is equidistant from all three sides of the triangle LMN.

(ii) From above part we get,

A is equidistant from MN and LN and also from LN and LM.

We get,

A is equidistant from MN and LM.

∴ A lies on angle bisector of ∠LMN.

Hence, proved that AM bisects ∠LMN.

Use ruler and compasses only for this question.

(i) Construct ∆ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.

(ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.

(iii) Construct the locus of points inside the triangle which are equidistant from B and C.

(iv) Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.

Answer

Steps of construction:

(i) Draw line BC = 6 cm and construct angle CBX = 60°. Cut off AB = 3.5 cm. Join AC, triangle ABC is the required triangle.

(ii) The locus of points which are equidistant from two sides is the angle bisector of angle between the lines. Construct BX, the bisector of angle B.

(iii) The locus of points inside the triangle which are equidistant from B and C is YZ, the perpendicular bisector of BC inside the triangle.

(iv) Let perpendicular bisector of BC meets BX at P.

Thus, P is the point which is equidistant from BA and BC, which is also equidistant from B and C. Measure PB.

Hence, PB = 3.5 cm