 Hello and welcome to the session. In this session, we will discuss constructions and the loci. In constructions and the loci, we find the point of intersection of two or more loci under given conditions. Here is my right to find which is equidistant from two points until which are 1.5 cm from the straight line be equal to 30 degrees. We can do this construction. Before that, let's find out the conditions given for the construction. We find the point A and this point A is given as equidistant from the two points P and Q and P and Q are 3 cm apart. So, this is one condition given to us. One condition given to us is that this point A is at a distance of 1.5 cm from the straight line QR where R is given as 40 degrees. So, the second condition is that the point A, a given distance to start with the construction and for this we find a certain construction we are given that the points P and Q are 3 cm apart. We draw PQ equal to 3 cm, PQ of measure 3 cm, measure 40 degrees. We make angle PQR equal to 40 degrees. So, this is angle PQR of measure 40 degrees. Now, let's see the first condition given to us. It says that the point A which we have to find out is equidistant from the MQ. We already know that the locus of a point equidistant from two given points is the right by sector of the straight line joining the two given points. According to the first condition the point A is equidistant from the two given points P and Q. So, to find the point A which satisfies this first condition we will draw the right by sector of PQ. So, the point A would lie on the right by sector of PQ. So, in the next step the right by sector of the line joining the two given points P and Q that is PQ. The right by sector of PQ, so the point A would lie on this right by sector XY. Now, the second condition given to us says that the point A is at a distance of 5 cm from this straight line QR. And we know that the locus of a point at a given distance from a fixed straight line is a pair of straight lines parallel to the fixed straight line at the given distance. Straight line as we have that the point A is at a distance of 1.5 cm from the straight line QR. So, we will draw a pair of straight lines parallel to QR which are at a distance of 1.5 cm from QR. Now, in the next step we draw 10DQ to X equal to QT equal to 1.5 cm. We put 10DQ to QR and we have cut off QT and QS equal to 1.5 cm on SQT. Draw a pair of straight lines parallel to QR at a distance of 1.5 cm from the points SMT equal to QR to QR and it is at a distance of 1.5 cm from QR. Dash is parallel to QR at a distance of 1.5 cm and let dash intersect XY the perpendicular bisector of PQ at the point A dash intersect XY. Then these points are the required points are the required points if the given conditions that is the point A or A dash is equidistant from the points P and Q of 1.5 cm from the straight line QR. Let us do another construction on which we need to OQ which are the arms of measure 60 degrees. These centimeters from the point Q all conditions are given to us by this construction we are given that the point A which we have to find out from OP and OQ. So, the first condition is OQ of measure 60 degrees because it is given to us at the point A in nose such that it is always 3 centimeters from the point Q. Now to do this construction we will follow the construction and we will draw the under P and Q of measure 60 degrees so P and Q of measure 60 degrees. P and Q of measure 60 degrees the point A is equidistant from P and Q that is the point A would be equidistant from OP and OQ that the local center point equidistant from straight lines is the bisector of the angle between the given intersecting straight lines. So, OP and OQ are two intersecting straight lines which are intersecting each other at this point O. We can start the angle bisector of angle POQ when the point A would lie on the angle bisector of angle POQ according to the first condition given to us. Then in the next step the bisect the angle is the angle bisector of angle POQ. If A would lie on the second condition we have that the point A is equidistant from the fixed point Q and we already know that the local center point equidistant from a fixed point is the second condition given with Q as center two meters. This circle taking Q as the center and three centimeters as the radius stands on the point Q and distance of three centimeters is this circle that we have drawn in the section of OX and this circle A is equidistant from OP OQ. Then this time A dash is also equidistant from OP and OQ and point A is equidistant from Q and point A dash is also equidistant from Q. Since we can say how we can do the constructions and conditions so this completes the session hope you have understood this concept.