 Hello and welcome to the session. In this session we will discuss basic locus theorem. First theorem that we have states the locus of a point equidistant from two fixed points is the right by sector of the straight line joining the fixed points. We can also say this as the right by sector locus. Let us now discuss the proof of this theorem. Consider case 1 in which we are given fixed points p and 2 and a moving point we taken as in the point of the theorem we have that this moving point x is equidistant from the two fixed points p and 2. So this means px would be equal to qx and in this case we will prove that the moving point x lies on the right by sector of the line joining the two fixed points that is pq. So consider these fixed points p and q and this x is a moving point we have to prove that x lies on the right by sector of pq for this. Let us do some construction join pq then bisect pq at y join so we have joined pq px qx xy and bisected pq at point y. Now let us carefully prove consider the triangles is equal to qx is given to us then y is equal to qy by the construction that we have done that is we have bisected pq at y so py would be equal to qy then yx is equal to yx which is a common site for both the triangles. So therefore we can say that the triangle p yx is congruent to the triangle by the SSS congruency criteria and so angle p yx would be equal to angle qyx as they are the corresponding parts of the congruent triangles so they are equal. Now from the figure we can see that angle p yx plus angle qyx is equal to 180 degrees as they form a linear pair and the line p yq is a straight line now since p yx and qyx angles are equal and also their sum is equal to 180 degrees so this means that angle p yx is equal to angle qyx and each would be equal to 90 degrees that is 180 degrees upon 2 which means that this angle is also of pressure 90 degrees and this angle is also of pressure 90 degrees therefore we can say that xy is the perpendicular to pq and also y is the midpoint of pq xy is perpendicular to pq and y is the midpoint of pq hence we can say that xy is the right by sector of pq that is x lies on the right by sector pq so this is what we were supposed to prove in this case so hence proved for case 1 now we consider the case 2 in this case we are given fixed points and q also we have as the right by sector and we have taken a point x on f t which is the right by sector of pq so here we have fixed points p and q is the right by sector of pq this point x we have taken on the right by sector of pq that is on f t and we have joined px and qx now we have to prove x is equal to qx this point b point y now to prove px equal to qx we consider the triangles p yx q yx in these two triangles we have angle p yx is equal to angle q yx and each is equal to 90 degrees as we know is the right by sector and we also have yx is equal to yx that is the common side of both the triangles and is equal to qy by sector of therefore triangle to the triangle qyx by the SAS congruency criteria now since these two triangles are congruent so we can say that px is equal to qx as they are the corresponding parts of the congruent triangles and so they are equal and this is what we were supposed to prove so hence we have proved the second case also so in case one we have proved now in case one we were given two fixed points and a moving point which was equidistant from the two fixed points and we have shown that the moving point lies on the right by sector and the line joining the fixed points thus we say from the case one let every point equidistant from points is on the right by sector pq and in the second case we were given two fixed points and the moving point on the right by sector of the line joining the fixed points and we have to prove that the moving point is equidistant from the two fixed points so from case two we can say that every point we write by sector of pq equidistant conclude that the right by sector straight line pq be required locus that is the locus of a point equidistant from two fixed points is the right by sector of the straight line joining the fixed points now let us discuss the next theorem which says that the locus of a point equidistant from two a pair of the angles between the two given straight lines and we can also say this as the angle by sector locus this theorem for this we first consider the case one in which we are given straight lines and they both intersect each other at point o and we have a point a which is the moving point that we take a n perpendicular to x y perpendicular to pq and also this point a is moving such that a n is equal to a m now in this case we need to prove that this moving point a lies on the angle by sector of angle x o q and we have drawn a n perpendicular to x y and a m perpendicular to pq and also a n is equal to a m we have to prove that this point a lies on the angle by sector of angle x o q let us first do some construction in this we join o a so we have produced o a to b now to start with the proof we consider the triangles o a n see these two triangles are right angle triangles so now in these two triangles is equal to o a which is the common hypotenuse for both the triangles then is equal to a m this is given to us and also angle n o is equal to angle a m o at each is equal to 90 degrees is perpendicular to x y is perpendicular to pq therefore from this we conclude the triangle o a n is congruent to the triangle by the RHS congruence criteria and thus we can say that angle a o n is equal to angle a o n that is these two angles are equal as per the corresponding parts of the congruent triangle so they are equal hence from this we can say that the moving point a lies on the bisector of angle x o q and this is what we were supposed to prove in case one let us consider the case two in this we are given straight lines each other at point o also o b angle x o q and we have a point a o b then is the perpendicular to x y and also a m is perpendicular to p q so we have two straight lines x y and p q we have o b as the angle bisector of angle x o q which means angle b o x and angle b o q are equal we have taken any point a on the o b which is the angle bisector of angle x o q and we have drawn a m perpendicular to x y a m perpendicular to p q now to prove in this case is equal to a m for this we consider the triangles o n a m a in these two triangles angle o n a is equal to angle o m a and each is equal to 90 degrees as we have a n is perpendicular to x y and a m is perpendicular to p q then the site o a is equal to site o a which is the common site to both the triangles and also angle a o n would be equal to angle a o m as we know that o b is angle x o q now from these three conditions we conclude that triangle o n a is congruent to the triangle o m a by a a s congruency criteria at these two triangles are congruent so from here we can conclude that a n is equal to a m as we have the corresponding parts of the congruent triangles and so they are equal this is what we were supposed to prove in case two so hence prove this in case one we were given two straight lines intersecting each other and we have taken a moving point equidistant from the two straight lines and we had shown that this moving point a lies on the angle bisector of angle x o q which is the angle formed between the two given straight lines now from case one we conclude that every point equidistant intersecting straight lines p q x y lies on the bisector of angle x o q and in case two we were given two intersecting straight lines