Locus

(i) False

The locus of the points situated at equal distances from a line is the lines parallel to the line on its both sides.

(ii) True

(iii) False

Three given points will be collinear only when they belong to the same set of numbers.

(iv) False

It is not always true as if the lines are parallel then it will be a line parallel to them while in case of intersecting lines the line will be the bisector of the angle formed at the point of intersection.

(v) True

Given here a quadrilateral of sides MN , NK, KL and ML

Also it is given that the diagonals of the quadrilateral bisect each other.

To prove: Quad KLMN is a parallelogram

Proof: As the diagonals bisect each other,

So, NO=OL and KO=OM……………..(1)

Now, in ∆KOL and ∆NOM,

∠KOL=∠NOM (vertically opposite angles)

KO=OM (from (1))

NO=OL (from (1))

∴ ∆KOL ∆NOM (by SAS rule)

So, KL=NM (by cpct)

∠OMN=∠OKL (by cpct)………..(2)

∠OLK=∠ONM (by cpct)…………(3)

Similarly, ∆KON ∆LOM (by SAS rule)

So, KN=LM (by cpct)

∠ONK=∠OLM (by cpct)………..(4)

∠OKN=∠OML (by cpct)…………(5)

From (2) and (3) ,

We can say they are alternate interior angles.

So, KL||NM

Similarly, From (4) and (5) ,

We can say they are alternate interior angles.

So, KN||ML

Hence, KLMN is a parallelogram.

The locus of a point equidistant from three non-collinear points A, B and C is the circumcentre of the triangle.

The locus of the points is the point of intersection of the perpendicular bisectors of each of the two points. The circumcentre is equidistant from all the three points A, B and C.

Example:

Here in ∆RST, U is the circumcentre of the triangle and RC ,SB and TA are the perpendicular bisectors of sides ST, RT and RS.

The locus of a point equidistant from three collinear points A, F and B is a null set that means there is no such locus of points.

Given that a circle with centre I and passing through points A and B

To prove: IC is the perpendicular bisector of AB

Proof: Here, IA=IB (radius)

So, ∆IAB is isosceles triangle.

So, ∠IAC=∠IBC………………..(1)

As we know that the altitude IC on AB is perpendicular bisector of AB.

Hence, ∠ICA=∠ICB=90°

And AC=BC

As ∆ICA and ∆ICB are congruent triangles.

Proved.

In the given figure ∆PBC and ∆QBC are isosceles triangles.

So, PB=PC and BQ=QC

To prove: line joining the points P and Q bisects line BC at right angles.

Proof: Now, in ∆PBQ and ∆PCQ

PB=PC (given)

BQ=QC (given)

PQ=PQ (common)

∴ ∆PBQ ∆PCQ (by SSS rule)

∠BPO=∠CPO (by cpct)………(1)

And ∠BQO=∠CQO (by cpct)…………(2)

Now in ∆BOP and ∆COP,

BP=CP (given)

OP=OP (common)

∠BPO=∠CPO (from (1))

∴ ∆BOP ∆COP (by SAS rule)

So, BO=CO (by cpct)

∠BOP=∠COP (by cpct)………(3)

Now, ∠BOP+∠COP=180°

⇒ ∠BOP+∠BOP=180° (from (3))

⇒ 2∠BOP = 180°

⇒ ∠BOP = 90°

Hence , ∠BPO=∠CPO=90°

∴ PQ bisects BC at right angles.

Given that ∆PQR and ∆SQR lie on the same side of the common base QR

Also, ∆PQR and ∆SQR are isosceles triangles .

So, QS=RS and QP=RP…………(1)

To prove : SP is the perpendicular bisector of line QR

Proof : In ∆PQS and ∆PRS,

QS=RS (given)

QP=RP (given)

SP=SP (common)

∴ ∆PQS ∆PRS (by SSS rule)

So, ∠PSQ=∠PSR (by cpct)…………….(2)

Now, in ∆QSO and ∆RSO,

QS=RS (given)

SO=SO (common)

∠PSQ=∠PSR (from (2))

∴ ∆QSO ∆RSO (by SAS rule)

So, QO=RO (by cpct)

And ∠QOS=∠ROS (by cpct)………..(3)

Now, ∠QOS+∠ROS=180° (straight angle)

⇒ ∠QOS+∠QOS=180° (from (3))

⇒ 2∠QOS =180°

⇒ ∠QOS=90°

Hence, SO is the perpendicular bisector of QR

So, SP is also the perpendicular bisector of QR

Given that PS is the bisector of ∠P intersects side QR at S.

Also, SN ┴ PQ and SM ┴PR

So, ∠NPS=∠MPS…………(1)

And ∠PNS=∠PMS ……………..(2)

In ∆PNS and ∆PMS

∠NPS+∠PNS+∠PSN=180°

And ∠MPS+∠PMS+∠PSM=180°

From (1) and (2)

180°-∠PSN=180°-∠PSM

⇒ ∠PSN=∠PSM………….(3)

Now, in ∆PNS and ∆PMS,

PS=PS (common)

∠NPS=∠MPS (given)

∠PSN=∠PSM (from (3))

∴ ∆PNS ∆PMS (by ASA rule)

So, NS=MS (by cpct)

Hence, proved.

Draw angle bisector BX of ∠ABC.

Take any point P on BX. Now draw ⊥ on AB and BC.

PL ⊥ AB and PM ⊥ BC

∴ ∠PLB = ∠PMB = 90°

In Δ PLB and Δ PMB,

∠PLB = ∠PMB (by construction)

∠LBP = ∠PBM (BP is bisector of ∠B)

BP = BP (common)

∴ Δ PLB ≅ Δ PMB (By AAS congruency)

⇒ PL = PM (By CPCT)

Thus, point P is the interior of ∠ABC and equidistant from AB and BC. So P is the locus.

The locus of point ℓ equidistant from three vertices is the circumcentre and three sides of a triangle is the incentre of the triangle.

Here, J is the incentre of the ∆LMN.

At first draw the angle bisectors of all the angles i.e, ∠L , ∠M and ∠N.

LO, MP and NQ are the angle bisectors respectively.

Let LO, MP and NQ intersect at J,

∴ J is the incentre.

Here, U is the circumcenter of the ∆RST.

At first , draw the perpendicular bisectors of the sides ST, RT and RS of the triangle.

Here, RC, SB and AT are the required perpendicular bisectors.

Lat them intersect at U

Hence, U is the circumcenter of the triangle.

Given that in ∆ABC , medians AD, BE and CF intersects at a point G.

AG = 6 cm, BE = 9 cm and GF = 4.5 cm

As we know that if the medians in a triangle intersect at a point then the point is the centroid of the triangle and the centroid divides the median in 2:1 ratio.

Here, G is the centroid of the ∆ABC.

And AG:GD = 2:1,

BG:GE = 2:1 ,

CG:GF = 2:1

Now,

GD=3 cm

Again,

⇒ CG=4.5×2=9

CG=9 cm

BE = BG+GE………………(1)

⇒BG=2GE

Putting in equation (1)

9 = 2GE+GE

⇒ 3GE=9

⇒ GE = 3

GE = 3 cm

Hence, GD=3 cm, CG=9 cm and GE = 3 cm

As we know that in a triangle, sum of any two sides is always greater than the third side.

So, in ∆AGB,

AG+GB>AB…………….(1)

Also, AD=AG+GD

And BE=BG+GE

As we know that if the medians in a triangle intersect at a point then the point is the centroid of the triangle and the centroid divides the median in 2:1 ratio.

Here, G is the centroid of the ∆ABC.

So, AG:GD=2:1 and BG:GE=2:1

⇒ AD=3GD…………………..(2)

And AG=2GD…………………(3)

Dividing equation (2) by (3)

⇒

……………….(4)

Again,

⇒ BE=3GE…………………..(2)

And BG=2GE…………………(3)

Dividing equation (2) by (3)

⇒

……………….(5)

Putting the value of AG and BG in equation (1)

AG+BG>AB

⇒

⇒

Proved.

Given a ∆ABC in which AD, BE and CF are the medians on the sides BC , AC and AB.

To prove: AD+BE>CF

BE+CF>AD

AD+CF>BE

Proof:

We will extend AD to H to form ∆BHC

Also, AG=GH………………(1)

F is the midpoint of AB and G is the midpoint of AH.

So, by midpoint theorem,

FG||BH

And FG= 1/2 BH

Similarly, GC||BH and BG||CH

So, we can see from the above that

BGCH is a parallelogram.

So, BH=GC …………(2)

And BG=HC…………..(3)

Now, in ∆BGH,

BG+GH>BH

⇒ BG+AG>GC (from (1),(2))

So, BE+AD>CF

Similarly, BE+CF>AD

AD+CF>BE

Proved.

Given that in ∆ABC , medians AD, BE and CF intersects at a point G.

As we know that the sum of any two sides is always greater than the third side in a triangle.

Here, G is the centroid of the ∆ABC

Now, in ∆ADB and ∆ADC,

AD+BD>AB………..(1)

AD+DC>AC…………(2)

In ∆BEC and ∆BEA,

BE+EC>BC………..(3)

BE+AE>AB………..(4)

In ∆CFA and ∆CFB,

CF+AF>AC………..(5)

CF+FB>BC………..(6)

Adding equation (1),(2),(3),(4),(5)and (6)

2AD+2BE+2CF+(BD+DC)+(EC+AE)+(AF+FB)>2AB+2BC+2AC

⇒ 2(AD+BE+CF)+BC+AC+AB>2(AB+BC+AC)

⇒ 2(AD+BE+CF)> 2(AB+BC+AC)-(BC+AC+AB)

⇒ 2(AD+BE+CF)> AB+BC+AC

Multiplying by 2 in the above equation

⇒ 4(AD+BE+CF)> 2(AB+BC+AC)

Hence, Proved.

Given that P is the orthocenter of ΔOBC.

To prove: the orthocenter of ΔABC is the point A

Proof:

P is the orthocenter of ΔABC

As we know that orthocenter is the point of all the perpendicular bisectors of the sides of a triangle.

Let AO is extended to D , BO is extended to E and CO is extended to F respectively.

So, AD=AO+OD

BE=BO+OE

And CF=CO+OF

As AD, BE and CF are the perpendicular bisectors

So, ADꞱBC , BEꞱAC and CFꞱAB

We can say that ADꞱBC,

ABꞱCO and

ACꞱBO

So, A is the orthocenter of ΔOBC.

Given that in ∆ABC, medians AD, BE and CF intersects at a point G.

GF = 4 cm and AD=7.5 cm

As we know that if the medians in a triangle intersect at a point then the point is the centroid of the triangle and the centroid divides the median in 2:1 ratio.

Here, G is the centroid of the ∆ABC.

And AG:GD = 2:1,

BG:GE = 2:1 ,

CG:GF = 2:1

Now, AD=AG+GD

⇒AG=2GD………….(1)

AD = AG+GD………………(1)

⇒AG=2GD

Putting in equation (1)

7.5 = 2GD+GD

⇒ 3GD=7.5

⇒ GD = 2.5

GD = 2.5 cm

Again,

⇒ CG=4×2=8

CG=8 cm

Hence, GD=2.5 cm, and CG=8 cm

For circumcentre we have to show AD is perpendicular bisector of BC.

In Δ ABD and Δ ADC,

AB = AC (given)

AD = AD (common)

BD = DC (D is midpoint of BC)

Δ ABD ≅ Δ ADC (BY SSS congruency)

⇒ ∠ ADB + ∠ ADC = 180°

⇒ AD ⊥ BC,

⇒ BD = DC

So AD is perpendicular bisector of BC>

So, the circumcentre lie on AD.

For incentre we have to show AD is bisector of ∠BAC.

Since Δ ABD ≅ Δ ADC

⇒ ∠BAD = ∠CAD ( By CPCT)

⇒ AD is the bisector of ∠BAC.

Hence, incenter lies on AD.

For orthocenter we need to prove AD is altitude corresponding to side BC.

Since Δ ABD ≅ Δ ADC

⇒ ∠ ADB + ∠ ADC = 180°

⇒ AD ⊥ BC,

⇒ AD is altitude corresponding to side BC.

For centroid we have to prove that AD is median corresponding to BC.

Since, it is given that D is the midpoint of BC. Ad is the median.

So, centroid lies on AD.

Hence Proved

Since , H is the orthocenter of the ΔABC

So, ADꞱBC , BEꞱAC and CFꞱAB

X, Y and Z are the midpoints of AH, BH and CH respectively

∴ XYZ is a triangle.

Hence, XZ||AC

So, ∠HOX=∠HEA=90°

Similarly, XY||AB

And ∠HPX=∠HFA=90°

And YZ||BC

So, ∠HQZ=∠HDC=90°

Hence, HO, HP and HQ are the altitudes of ∆XYZ

So, H is the orthocenter of ∆XYZ

In Δ ABC, draw bisector of BC which cuts it at D.

Take any point P on AD. Draw PN⊥ AB and PM ⊥ AC.

In Δ APN and Δ APM

∠ PNA = ∠ PMA = 90° ( by construction)

∠PAN = ∠PAM ( AD is bisector of ∠A)

AP = AP )( common)

Δ APN ≅ Δ APM (By AAS rule)

⇒ PN = PM ( By CPCT)

So, P is equidistant from AB and AC.

∴ any point on AD will be equidistant from AB and AC.

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The locus of the end of the pendulum of the clock is an arc of the circle.

Construction: Join DE and DF.

In ΔABC, D and E are the mid points of BC and AC.

∴ DE || AB

DE = 1/2 AB

Now DE||AB

⇒ FA||DE ….. (1)

Similarly EA||DF ….. (2)

From (1) and (2),

EAFD is parallelogram.

As diagonals of ||gm bisects each other,

EF will bisect AD.