 Hello and welcome to the session. In this session first we shall discuss converse problem for section formula. Let us quickly revise section formula first. The section formula states that the coordinates of the point which divides internally the line joining two given points in a given ratio is given by x is equal to n1 into x2 plus n2 into x1 upon n1 plus n2 and y is equal to n1 into y2 plus n2 into y1 whole upon n1 plus n2 here p is the point with the coordinates xy which divides internally the line joining the two given points a and b with the coordinates x1 by 1 and x2 by 2 respectively in the given ratio that is n1 is to n2 If the division is external then the coordinates xy of point p is given by x is equal to n1 into x2 minus of n2 into x1 whole upon n1 minus n2 and y is equal to n1 into y2 minus of n2 into y1 whole upon n1 minus n2 For external division the coordinates xy of point p which divides externally the line joining the two given points a and b with the coordinates x1 y1 and x2 y2 respectively in the given ratio that is n1 is to n2 is given by x is equal to n1 into x2 minus n2 into x1 whole upon n1 minus n2 and y is equal to n1 into y2 minus n2 into y1 whole upon n1 minus n2 Now we shall discuss the converse of what we had just studied known as converse problem for section formula Converse problem is to find out the ratio in which a given point divides the join of the given points Let us take an example if the point p whose coordinates are 4 to divides the line segment joining the points a and b whose coordinates are 2 minus 4 and 814 respectively Find the ratio in which point p divides a and b Here we are given a point p whose coordinates are 4 to which divides the line segment joining the points a and b with the coordinates 2 minus 4 and 814 respectively Now we need to find the ratio in which point p divides a and b Here we are given point a with the coordinates 2 minus 4 point b with the coordinates 814 and point p with the coordinates 4 to So the value of x1 is equal to 2 the value of y1 is equal to minus 4 x2 is equal to 8 and y2 is equal to 14 The value of x is given as 4 and the value of y is given as 2 Now let the ratio be mu is to 1 that is n1 is equal to mu and n2 is equal to 1 and we need to find this ratio Now the working rule is as follows The first step is let the required ratio be mu is to 1 Now using section formula we know that x is equal to n1 into x2 plus n2 into x1 whole upon n1 plus n2 and y is equal to n1 into y2 plus n2 into y1 whole upon n1 plus n2 Now we shall calculate coordinates of point p which divides the given points a and b in the ratio mu is to 1 Then coordinates of point p are given by x is equal to n1 into x2 that is mu into 8 plus n2 into x1 that is 1 into 2 that is 2 whole upon n1 plus n2 that is mu plus 1 and y is equal to n1 into y2 that is mu into 14 plus n2 into y1 that is 1 into minus 4 whole upon n1 plus n2 that is mu plus 1 which implies that x is equal to 8 mu plus 2 upon mu plus 1 and y is equal to 14 mu minus 4 upon mu plus 1 Now in the next step we equate the coordinates obtained in step 2 to the given coordinates that is equating the x coordinates we get x is equal to 8 mu plus 2 upon mu plus 1 which is equal to 4 which implies that 8 mu plus 2 is equal to 4 into mu plus 1 that is 4 into mu plus 4 which implies that 8 mu minus 4 mu that is equal to 4 mu is equal to 4 minus 2 that is 2 therefore mu is equal to 2 by 4 which is equal to 1 by 2 or we can also equate the y coordinates that is equating the y coordinates we get y is equal to 14 mu minus 4 upon mu plus 1 which is equal to 2 which implies that 14 mu minus 4 is equal to 2 into mu plus 1 that is 2 mu plus 2 which implies that 14 mu minus 2 mu is equal to 12 mu which is equal to 2 plus 4 that is 6 therefore the value of mu is equal to 6 by 12 which is equal to 1 by 2 therefore from both the cases we can say that the required ratio is minus 2 to Since the value of mu is primitive therefore the division is internal if mu comes out to be negative then the division will be external now we shall discuss how to find area of a co-relation let A B C D be a co-relation and area is to be determined A B C D is a co-relator whose area is to be determined we divided into two triangles triangle A B C and triangle A C D now we find the area of these triangles and by adding the area of both the triangles we get the area of the co-relator A B C D for calculating this area we take the vertices in order that is point A with the coordinates A 1, A 2 point B with the coordinates B 1, B 2 point C with the coordinates C 1, C 2 and point D with the coordinates D 1, D 2 area of co-relator A B C D is equal to area of triangle A B C plus area of triangle A C D now area of triangle A B C is equal to 1 by 2 into A 1 into B 2 minus of A 2 into B 1 plus B 1 into C 2 minus of C 1 into B 2 plus C 1 into A 2 minus of C 2 into A 1 plus area of triangle A C D which is given by 1 by 2 into A 1 into C 2 minus of A 2 into C 1 plus C 1 into B 2 minus of C 2 into B 1 plus D 1 into A 2 minus of A 1 into B 2 on solving this we get 1 by 2 into A 1 into B 2 minus of A 2 into B 1 plus B 1 into C 2 minus of B 2 into C 1 plus C 1 into B 2 minus of C 2 into B 1 plus D 1 into A 2 minus of A 1 into B 2 which can also be written in this form that is we write the x coordinates and y coordinates of the vertices in two columns as shown the x coordinate and y coordinate of the first vertex is occurring twice and we multiply diagonally attaching negative sign to the multiplication from right to left and positive sign to multiplication from left to right and divide the result by 2 let us take an example find the area of the quadrilateral whose vertices are 2 4 minus 1 minus 2 3 minus 5 and 6 1 let A B C D be the given quadrilateral with vertices A with the coordinates 2 4 B with the coordinates minus 1 minus 2 C with the coordinates 3 minus 5 and D with the coordinates 6 1 and we know that area of quadrilateral A B C D is given by 1 by 2 into 2 into minus 2 minus of 4 into minus 1 plus minus 1 into minus 5 minus of minus 2 into 3 plus 3 into 1 minus of minus 5 into 6 plus 6 into 4 minus of 1 into 2 which is equal to 1 by 2 into 2 into minus 2 that is minus 4 minus 4 into minus 1 that is plus 4 plus minus 1 into minus 5 that is 5 minus of minus 2 into 3 that is plus 6 plus 3 into 1 that is 3 minus of minus 5 into 6 that is plus 30 plus 6 into 4 that is 24 minus 1 into 2 that is minus 2 therefore we have 1 by 2 into minus 4 plus 4 that is 0 plus 5 plus 6 that is 11 plus 3 plus 30 that is 33 plus 20 per minus 2 that is 22 therefore we have 1 by 2 into 11 plus 33 plus 22 that is 66 which is equal to 33 therefore area of the given quadrilateral 33 square units now we are going to discuss slope of a line making equal angles with the axis here we shall discuss 2 cases 1 when the value of theta is equal to 45 degrees and the second case when the value of theta is equal to 145 degrees now our first case was when the value of theta is equal to 45 degrees then the value of the slope that is m is equal to tan of theta that is tan of 45 degrees which is equal to 1 and in the second case we have the value of theta as 145 degrees where the value of the slope that is m is equal to tan of theta that is tan of 135 degrees which is equal to minus 1 therefore the value of m is equal to plus minus of 1 when the line makes equal angle with the axis we should also note that when a line make equal angles with the axis it cuts off equal intercepts on the axis that is when a line makes equal angles with the axis it cuts off equal intercepts on the axis also we have if a line is parallel to the x axis its integration is 0 degrees that is tan of theta which means tan of 0 degrees is equal to 0 and if a line is perpendicular to the x axis its inclination is 90 degrees its slope that is tan of theta which is equal to tan of 90 degrees is equal to infinity that is a line parallel to y axis has no slope as tan of 90 degrees does not exist this completes our session hope you enjoyed this session