The document contains solutions to 8 questions about locus and geometry. The key points are: (1) The locus of points equidistant from two lines is the line perpendicular to both lines. (2) If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. (3) The locus of points equidistant from three non-collinear points is a circle centered at their intersection point.

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Rajasthan Board RBSE Class 10 Maths Chapter
10 Locus Ex 10.1
Question 1.
Write true of false for following statements and also give reason of your answer.
(i) Set of points at same distant from a line is a line.
(ii) A circle is a locus of those points which lie at constant distant from given point.
(iii) Three given points will be collinear when they are not the element of the set of points of line.
(iv) The locus of a point, which is equidistant from two lines will be line parallel to both the lines.
(v) The locus of a point which is equidistant from two fixed points is the perpendicular bisector of
the line joining the two fixed points.
Solution :
(i) False : Locus of points at equal distance from a line are lines parallel to that line in both sides.
(ii) True : The fixed point is center and fixed distance is radius.
(iii) False : Three given points will be collinear when all three points lie on same line such that three
points are elements of set of points of that line.
(iv) False : It depends on the position of lines. If both are parallel then ¡t will be parallel and if
intersecting lines then line will be angular bisector of intersecting points.
(v) True.
Question 2.
The diagonals of a quadrilateral bisect each other. Prove that this quadrilateral is parallelogram.
Solution :
Given, A quadrilateral ABCD whose diagonals AC and BD bisects at point O.
i.e., OA = OC and OB = OD.
To Prove : ABCD is a parallelogram.
Proof : ΔAOB and ΔCOD
OA = OC (Given)
∠AOB = ∠COD (Vertically opposite angles)
and OB = OD (given)
ΔAOB ≅ ΔCOD (By ASA rule)
⇒ ∠OAB = ∠OCD (CPCT)
⇒ ∠CAB = ∠ACD
But these are alternate angles made by transversal at lines AB and CD. Hence AB || CD.
Similarly, AD || BC
Hence, ABCD is a parallelogram.
Question 3.
What will be the locus of points equally distant from three non-collinear points. A, B and C. Justify
your answer.
Solution :
The locus of points will be center of circle.
Justification : Let A, B and C are three non-collinear points. Join AB and BC.

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Let P is a moving point which is equidistant from A and B i.e. P is perpendicular bisector of AB.
Again, P is equally distant from B and C, so P is perpendicular bisector of BC.
So, P is intersecting point of perpendicular bisectors AB and BC.
Thus, P is the center of the circle passing through three given points.
Question 4.
What will be the locus of points equally distant from three collinear points. Clarify your answer.
Solution :
Let three distinct points A, B and C lie on line C.
Any point at equally distant from points A and B lie on perpendicular bisector of AB. Thus that point
lie on line m which is perpendicular bisector of AB.
Similarly, any point equally distant from points B and C and lie on line n which is perpendicular
bisector of BC.
Similarly, any point at equally distant from points A, B and C and common perpendicular of two
lines m and n.
but m ⊥ AB and n ⊥ BC
⇒ m ⊥ AC and n ⊥ AC ⇒ m || n
∴ This is no common point in m and n. Therefore, there is no point which is equidistant from three
linear points.
Hence these points dd not exist.
Question 5.
Prove that locus of centers of circles passing through points A and B is perpendicular bisector of line
segment AB.
Solution :
Let P and Q be the centers of two circle C and C, each passing through two given points A and B.
Then,
PA = PB (radii of the circle C)
⇒ P lies on the perpendicular bisector of AB ….(i)
Again QA = QB (radii of the circle C)
⇒ Q lies on the perpendicular bisector of AB ……(ii)
From (i) and (ii) it follows that P and Q both. lie on the perpendicular bisector of AB.
Hence, the locus of the centers of all the circles passing through A and B is the perpendicular bisector

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of AB.
Question 6.
In given figure, at common base BC, two isosceles triangles ∆PBC and ∆QBC lie on both sides of
BC. Prove that line joining P and Q bisects line BC at 90°.
Solution :
Given : ∆PBC and ∆QBC are two isosceles triangles which lie on both sides of base BC.
Here, PB = PC and BQ = CQ
To Prove : ∠POB = ∠POC = 90°
or ∠QOB = ∠QOC = 90°
Proof : In ∆PBC,
PB = PC (Given)
∴ ∠PBO = ∠PCO (Equal sides)
PO = PO (Common)
by S.A.S. congruence of
∆POB ≅ ∆POC (Corresponding sides of corresponding triangles are equal)
⇒ ∠PBO = ∠POC ….(i)
We Know that
∠PBO + ∠POC = 180°
∠PBO + ∠POB = 180° [From equation (i)]
2∠POB = 180°
∠POB = = 90°
∠PBO = ∠POC = 90°
Similarly, ∠QOB = ∠QOC = 90°
Thus, PQ, bisects BC at 90°
Question 7.
In given figure, two isosceles triangles PQR and SQR are formed on same side of phase QR. Prove
that line SP is perpendicular bisector of line QR.

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Solution :
Given : Two isosceles triangles PQR and SQR having same base QR and SQ = SR and PQ = PR
To Prove : SP is perpendicular bisector of QR.
Proof : ∵∆PQR is an isosceles triangle.
∠PQ = PR
Now ∠PQR = ∠PRQ
In isosceles triangle, angles opposite to equal sides are same.
In ∆PQM and ∆PRM,
PQ = PR (Given)
∠PQR = ∠PRQ
PM = PM (Common)
∴ ∆PQM ≅ ∆PRM
∴ ∠QMP = ∠RMP and QM = MR
Now, ∠QMP + ∠RMP = 180°
∠QMP = ∠RMP = 90°
Thus. PM ⊥ QR
∴ M is mid-point of QR.
Similarly perpendicular bisector of QR is PM.
Now point P lies on SM.
∵ ∆SQR is an isosceles triangle.
Perpendicular bisector of QR is SM.
Therefore line segment SP will be perpendicular bisector of base QR.
Question 8.
In given figure, PS, bisector of ∠P intersects side QR at point S. SN ⊥ PQ and SM ⊥ PQ are drawn.
Prove that SN = SM.
Solution :
Given : In ∆PQS, PS is bisector of ∠P and SN ⊥ PQ and SM ⊥ PR.
To Prove : SN = SM
Proof : In ∆PSN and ∆PSM
PS = PS (Common)
∠PNS = ∠PMS (Each 90° because SN ⊥ PQ and SM ⊥ PR)
∠NPS = ∠MPS (Since PS, bisector of ∠P)
by A.A.S. congruence rule
∆ASP ≅ ∆PSM

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∴ SN = SM
Corresponding sides of congruent angles are equal.
Question 9.
In the adjoining figure ∠ABC is given. Find the locus of a point in the interior of ∠ABC and
equidistant from BA and BC.
Solution :
Draw angular bisector BX of ∠ABC. Take any point P on BX. Now from P draw perpendicular on
AB and BC.
PL ⊥ AB and PM ⊥ BC
∴ ∠PLB = ∠PMB = 90°
∆PLB and ∆PMB,
∠PLB = ∠PMB (By construction)
∠LBP = ∠PBM (BP is bisector of ∠B)
BP = BP (Common)
∴ ∆PLB ≅ ∆PMB
By A.A.S. congruence rule,
PL = PM (In congruent triangles corresponding sides are equal)
Thus, point P is interior of ∠ABC and equidistant from AB and BC is required locus.