Question

Prove that The locus of a point
equidistant from two intersecting lines is the pair
of bisectors of the angles formed of these lines.
Fromn
The bisogtong of
ID​

Answer 1

Answer:

Step I : We initially prove that any point equidistant from two given intersecting lines lies on one of of the lies bisecting the angles fromed by the given lines. <br> Given :

and

are two line intersecting a O. P is the point on the plane such that PM = PN <br> Line I is the bisector of

and

<br> LIne m is the bisector of

and

<br> To proves : P lies either on the line I or on the line m <br> Proof : In

and

, PM = PN <br> OP is ommon side and

<br>

By RHS congrunce property,

<br> So,

, i.e P lies on the angles bisector of

<br> As l is the bisector of

and

, P lies on the line l. <br> Similarly, if P lies in any of the regions of

or

, such that it is equidistant from

and

then we can conclude that P lies on the angle bisector l or on the angle bisector m. <br> Step 2 : Now, we prove that any two point on the bisector of one of the angle formed by two intersecting lines is equidistant from the lines. <br> Given

and

, intersect at O lines l and m are the angle bisectors. <br> proof : Let l be the angle bisector of

and

and m be the angle bisector of

and

<br> Let P be a point on the angle bisector l, as shown in <br> If P coinclides with O, then P is equidistant line

and

<br> Suppose P is different O <br> Draw the perpendicular

and

from the point P onto the lines

and

respectively. <br> Then in

and OP is a common side. <br>

By the AAS congruance property. <br>

<br> So, PN = PM (

The corresponding sides in congruent triangles. <br> That is P is equidistant from lies

and

<br> Hence, from Step I and Step II of the proof, it can be said that the locus of the point, which is equidistant from the two intersecting lines is the pari of the angle bisectors of the two pairs of vertically oppoiste angles formed by the lines.