 and welcome to the session. Today we will learn about coordinate geometry. Let us start with distance formula. Suppose we are given two points A and B where the coordinates of A are x1, y1 and the coordinates of B are x2, y2 and we need to find the distance between AB then AB will be equal to square root of x2 minus x1 whole square plus y2 minus y1 whole square. Let's see one example for this. Suppose we are given two points A with coordinates 7 comma minus 4 and B with coordinates minus 5 comma 1 and we need to find the distance between A and B then AB will be equal to square root of here this will be x1 y1 and this will be x2 y2. So AB will be equal to x2 minus x1 that is minus 5 minus 7 whole square plus y2 minus y1 whole square that is 1 minus minus 4 whole square this will be equal to square root of minus 12 whole square plus 1 plus 4 whole square which will be equal to square root of 144 plus 25 that is square root of 169 which will be equal to 13 therefore AB is equal to 13 units. Now suppose we have two points O with coordinates 0 comma 0 and P with coordinates x comma y and we need to find the distance OP then OP will be equal to square root of x minus 0 whole square plus y minus 0 whole square and this will be equal to square root of x square plus y square. Now the point O with coordinates 0 comma 0 is the origin and P is any point with coordinates x comma y. So the distance of a point P with coordinates x comma y from the origin is square root of x square plus y square. Now let's see section formula pose we are given a line segment PQ where the coordinates of PR x1 comma y1 and the coordinates of QR x2 comma y2 and there is a point A which divides the line segment PQ in the ratio m1 is to m2 and we need to find the coordinates of the point A. So let us suppose that the coordinates of AR x comma y then x will be equal to m1 into x2 plus m2 into x1 upon m1 plus m2 and y will be equal to m1 into y2 plus m2 into y1 upon m1 plus m2. Here PA is to AQ is m1 is to m2 and the point A divides the line segment PQ internally. Here is one example for this we are given the line segment PQ where the coordinates of PR minus 5 comma 11 and the coordinates of QR 4 comma minus 7 and the point A divides the line segment PQ in the ratio 7 is to 2. So let us suppose that the coordinates of AR x comma y so here this is x1 y1 this is x2 y2 and this is m1 m2 thus x will be equal to m1 x2 that is 7 into 4 plus m2 x1 that is 2 into minus 5 upon m1 plus m2 that is 7 plus 2 so this will be equal to 2 and y is equal to m1 y2 that is 7 into minus 7 plus m2 y1 that is 2 into 11 upon m1 plus m2 that is 7 plus 2 so this is equal to minus 3 and thus coordinates of AR 2 comma minus 3. Now suppose we are given a line segment PQ where the coordinates of PR x1 y1 and the coordinates of Q1 x2 y2 and we need to find the coordinates of the midpoint of PQ so here A is the midpoint of PQ and let us say that the coordinates of AR x comma y. Now as A is the midpoint so that means point A will divide the line segment PQ in the ratio 1 is to 1 so here PA is to AQ is 1 is to 1 so by section formula x will be equal to 1 into x2 plus 1 into x1 upon 1 plus 1 so this will be equal to x1 plus x2 upon 2 and y will be equal to 1 into y2 plus 1 into y1 upon 1 plus 1 that is y1 plus y2 upon 2 so the midpoint of the line segment joining the points P with coordinates x1 comma y1 and Q with coordinates x2 comma y2 is x1 plus x2 upon 2 comma y1 plus y2 upon 2 so here if you want to find out the midpoint of line segment PQ then the coordinates of A will be minus 5 plus 4 upon 2 comma 11 plus minus 7 upon 2 that will be equal to minus 0.5 comma 2 so the coordinates of AR minus 0.5 comma 2 now let's move on to area of a triangle suppose we are given a triangle ABC where the coordinates of AR x1 comma y1 coordinates of VR x2 comma y2 and coordinates of CR x3 comma y3 then the area of triangle ABC will be equal to 1 by 2 into x1 into y2 minus y3 plus x2 into y3 minus y1 plus x3 into y1 minus y2 let's see one example for this suppose we are given 3 points A with coordinates 1 comma 2 B with coordinates minus 2 comma 3 and C with coordinates minus 3 comma minus 4 and we need to find the area of triangle ABC now here this will be x1 y1 x2 y2 x3 y3 so area of triangle ABC will be 1 by 2 into x1 that is 1 into y2 minus y3 that is 3 minus minus 4 plus x2 that is minus 2 into y3 minus y1 that is minus 4 minus 2 plus x3 that is minus 3 into y1 minus y2 that is 2 minus 3 which will be equal to 1 by 2 into 1 into 3 plus 4 plus minus 2 into minus 6 plus minus 3 into minus 1 that is 1 by 2 into 7 plus 12 plus 3 that is equal to 11 therefore area of triangle ABC is equal to 11 square units now suppose we are given 3 points A V and C then they will form a triangle like this and now if we are given 3 points P Q R such that these three points are in a line then they will not form any triangle or we can say that area of triangle P Q R will be equal to 0 as there is no triangle now the points which are in a line are called collinear points so points P Q R are collinear if and only if area of triangle P Q R is equal to 0 so if you want to prove that three points are collinear then we should prove that the area of the triangle formed by these three points is equal to 0 so in this session we have learned distance formula the distance of a point from origin section formula how to find out the midpoint of a line segment area of a triangle and the condition for three points to be collinear with this we finish this session hope you must have understood all the concepts goodbye take care and have a nice day