(i) $\quad \mathrm{A}(9,3)$ and $\mathrm{B}(15,11)$ The given points are $A(9,3)$ and $B(15,11)$. Then $\left(x_{1}=9, y_{1}=3\right)$ and $\left(x_{2}=15, y_{2}=11\right)$ $A...

### Use ruler and compasses only for this question. (i) Construct the locus of points inside the triangle which are equidistant from B and C. (ii) Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.

Steps of construction: Draw line \[BC\text{ }=\text{ }6\text{ }cm\] and construct angle \[CBX\text{ }=\text{ }{{60}^{o}}\]. Cut off \[AB\text{ }=\text{ }3.5\]. Join \[AC\], triangle \[ABC\]is the...

### Use ruler and compasses only for this question. (i) Construct ∆ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60degree (ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.

Steps of construction: (i) Draw line \[BC\text{ }=\text{ }6\text{ }cm\] and construct angle \[CBX\text{ }=\text{ }{{60}^{o}}\]. Cut off \[AB\text{ }=\text{ }3.5\]. Join \[AC\], triangle \[ABC\]is...

### 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.

Construction: \[Join\text{ }AM\] Proof: (i) Since,\[A\] lies on bisector of \[\angle N\] So, \[A\]is equidistant from \[MN\text{ }and\text{ }LN.\] Now again, as \[A\] lies on the bisector...

### 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.

Construction: From \[P\], draw \[PL\bot AB~and~PM\bot BC\] Proof: In \[\vartriangle PLB\text{ }and\text{ }\vartriangle PMB\], \[\angle PLB\text{ }=\angle PMB\text{ }[Each\text{ }{{90}^{o}}]\]...

### Construct a triangle ABC in which angle ABC = 75o, AB = 5cm 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.

Steps of Construction: i) Draw a line segment \[BC\text{ }=\text{ }6.4\text{ }cm\] ii) At\[B\], draw a ray \[BX\]making an angle of \[{{75}^{o}}\] with \[BC\]and cut off \[BA\text{ }=\text{ }5\text{...

### Construct a right angled triangle PQR, in which ∠Q = 90degree, hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meet PR at point T. Prove that T is equidistant from PQ and QR.

Steps of Construction: i) Draw a line segment \[QR\text{ }=\text{ }4.5\text{ }cm\] ii) At \[Q,\]draw a ray \[QX\]of an angle of \[{{90}^{o}}\] iii) With\[centre\text{ }R\text{ }and\text{...

### In each of the given figures: PA = PB and QA = QB Prove in each case, that PQ (produced, if required) is perpendicular bisector of AB. Hence, state the locus of the points equidistant from two given fixed points.

(i) (ii) SOLUTION: Construction: Join \[PQ\]which meets \[AB\text{ }in\text{ }D.\] Proof: \[As\text{ }P\text{ }is\text{ }equidistant\text{ }from\text{ }A\text{ }and\text{ }B.\] So, \[P\]lies on the...

### 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.

Given: In triangle ABC, \[AB\text{ }=\text{ }4.2\text{ }cm,\text{ }BC\text{ }=\text{ }6.3\text{ }cm\text{ }and\text{ }AC\text{ }=\text{ }5\text{ }cm\] Steps of Construction: i) Draw a line segment...

### 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

solution: Draw \[XL\bot AC\text{ }and\text{ }XM\bot AB.\] Now, join \[YC\] Proof: (i) In \[\vartriangle AXL\text{ }and\text{ }\vartriangle AXM,\] \[\angle XAL\text{ }=\text{ }\angle XAM\text{...

### Given: CP is the bisector of angle C of ∆ABC. Prove: P is equidistant from AC and BC.

solution: Construction: \[From\text{ }P,\text{ }draw\text{ }PL\bot AC\text{ }and\text{ }PM\bot CB\] Proof: In \[\vartriangle LPC\text{ }and\text{ }\vartriangle MPC,\] \[\angle PLC\text{ }=\text{...

### Given: PQ is a perpendicular bisector of side AB of the triangle ABC. Prove: Q is equidistant from A and B.

solution: Construction: Join AQ Proof: In \[\vartriangle AQP\text{ }and\text{ }\vartriangle BQP,\] \[AP\text{ }=\text{ }BP\text{ }\left[ Given \right]\] \[\angle QPA\text{ }=\text{ }\angle QPB\text{...