 Hello and welcome to the session. In this session we are going to discuss the following question and the question says that proof or disprove that a figure defined by the 4 points 1, 6, 1 minus 3 minus 8 minus 3 and minus 8, 6 is a square. Let us start with the solution of the given question. In this question we are given 4 points and these points are 1, 6, 1 minus 3 minus 8 minus 3 and minus 8, 6. We want to prove or disprove that a figure defined by 4 given points is a square. For this we need to prove 4 conditions and these are Diagonals bisect each other. Length of Diagonals is equal. Length of all sides is equal and Diagonals bisect at 90 degrees. Now let us draw these points on the plane. Now here we have drawn these points on the plane and now we will join these points that is we join point A to point B, point B to point C, point C to point D and point D to point A. So this is the quadrilateral with vertices A, B, C and D. Here we see that A, C and B, D are the 2 Diagonals. The first condition we need to prove is Diagonals bisect each other. So to prove first condition we find midpoints of Diagonal A, C and Diagonal B, D. Now we shall find midpoint of Diagonal A, C. We know that midpoint of a line segment joining 2 points with coordinates x1, y1 and x2, y2 is given by the ordered pair x1 plus x2 whole upon 2, y1 plus y2 whole upon 2. So here we have point A with coordinates 1, 6. Let this be equal to the ordered pair x1, y1 and point C with coordinates minus 8, minus 3 and let this be equal to the ordered pair x2, y2. So midpoint of Diagonal A, C will be given by the ordered pair x1 plus x2 whole upon 2 that is 1 plus of minus 8 that is 1 minus 8 whole upon 2, y1 plus y2 whole upon 2 that is 6 plus of minus 3 which is 6 minus 3 whole upon 2. And this is equal to the ordered pair minus 7 by 2, 3 by 2. Also we shall find midpoint of Diagonal B, D. So midpoint of Diagonal B, D is given by the ordered pair x1 plus x2 whole upon 2 that is equal to 1 plus of minus 8 that is 1 minus 8 whole upon 2. y1 plus y2 whole upon 2 that is minus 3 plus 6 whole upon 2 and this is equal to the ordered pair minus 7 by 2, 3 by 2. Therefore, coordinates of midpoint of Diagonal A, C which is equal to the ordered pair minus 7 by 2, 3 by 2 is same as coordinates of midpoint of Diagonal B, D. So the coordinates of the midpoints are same thus we say that Diagonals bisect each other. So the first condition is proved. Now the second condition is length of Diagonals is equal. So we will find the length of Diagonals A, C and B, D. For this we will use distance formula that is distance between any two points with coordinates x1, y1 and x2, y2 is given by the formula that is square root of x2 minus x1 whole square plus y2 minus y1 whole square. So length of Diagonal A, C will be given by square root of x2 minus x1 whole square that is minus 8 minus 1 whole square plus y2 minus y1 whole square that is minus 3 minus 6 whole square and this is equal to square root of minus 8 minus 1 that is minus 9 whole square plus minus 3 minus 6 is minus 9 whole square and this is equal to square root of minus 9 whole square is 81 plus minus 9 whole square that is 81. So we have square root of 81 plus 81 which is equal to square root of 162 and that is equal to 9 into square root of 2. Now length of Diagonal B, D will be given by now let us take the coordinates of B that is 1 minus 3 as x1, y1 and coordinates of D that is minus 8, 6 as x2, y2. So length of Diagonal B, D will be equal to square root of x2 minus x1 whole square that is minus 8 minus 1 whole square plus y2 minus y1 whole square that is 6 minus of minus 3 whole square and this is equal to square root of minus 8 minus 1 is minus 9 whole square plus 6 minus of minus 3 that is 6 plus 3 which is equal to 9 square which is equal to square root of 81 plus 81 that is equal to square root of 162 which can be written as 9 into square root of 2. So here we have length of Diagonal A, C is equal to length of Diagonal B, D which is equal to 9 into square root of 2. So length of Diagonals is equal thus the second condition is also satisfied. Now we prove the third condition that is length of all sides should be equal. Now let us find length of side AB, side BC, side CD and side DA using distance formula. So length of side AB is given by square root of x2 minus x1 whole square that is 1 minus 1 whole square plus y2 minus y1 whole square that is minus 3 minus 6 whole square which is equal to square root of 1 minus 1 that is 0 square plus minus 3 minus 6 that is minus 9 whole square which is equal to square root of 0 plus 81 that is equal to square root of 81 which is equal to 9. Length of side BC is equal to square root of x2 minus x1 whole square that is minus 8 minus 1 whole square plus y2 minus y1 whole square that is minus 3 minus of minus 3 whole square and this is equal to square root of minus 8 minus 1 that is minus 9 whole square plus minus 3 minus of minus 3 is minus 3 plus 3 that is equal to 0 square and this is equal to square root of minus 9 whole square that is 81 plus 0 square that is 0 which is equal to square root of 81 which is equal to 9. Similarly, length of side CD is equal to square root of x2 minus x1 whole square that is minus 8 minus of minus 8 whole square plus y2 minus y1 whole square that is 6 minus of minus 3 whole square and this is equal to square root of minus 8 minus of minus 8 that is minus 8 plus 8 whole square plus 6 minus of minus 3 that is 6 plus 3 whole square and this is equal to square root of minus 8 plus 8 is 0 square plus 6 plus 3 whole square that is 9 square and this is equal to square root of 0 plus 81 that is equal to square root of 81 which is equal to 9 and lastly length of side DA is equal to square root of x2 minus x1 whole square that is 1 minus of minus 8 whole square plus y2 minus y1 whole square that is 6 minus 6 whole square and this is equal to square root of 1 minus of minus 8 that is 1 plus 8 whole square plus 6 minus 6 whole square that is 0 square and this is equal to square root of 1 plus 8 whole square that is 9 square plus 0 square that is 0 and this is equal to square root of 9 square plus 0 that is 81 plus 0 which is equal to 81 and this is equal to 9 thus we have seen that length of side AB is equal to 9 length of side BC is equal to 9 length of side CD is equal to 9 and length of side DA is also equal to 9 so we say that length of side AB is equal to length of side CD is equal to length of side BC and is equal to length of side DA which is equal to 9 thus we say that length of all sides are equal so the third condition is also satisfied now we prove the fourth condition we have to prove Diagnose AC and BD intersect at 90 degrees for this we will see the slope of the two diagonals we know that if the product of the slopes of two lines is minus 1 then the two lines intersect at right angle or 90 degrees we also know that slope of the line joining two points with coordinates x1, y1 and x2, y2 is given by y2 minus y1 the whole whole upon x2 minus x1 so now we find slope of Diagnose AC it is denoted by m1 and it is given by y2 minus y1 that is minus 3 minus 6 whole upon x2 minus x1 that is minus 8 minus 1 and this is equal to minus 9 upon minus 9 that is equal to 1 similarly slope of Diagnose BD let us denote it by m2 and this is equal to y2 minus y1 that is 6 minus of minus 3 whole upon x2 minus x1 that is minus 8 minus 1 and this is equal to 6 plus 3 whole upon minus 9 that is equal to 9 upon minus 9 which is equal to minus 1 so we have m1 is equal to 1 and m2 is equal to minus 1 so product of slopes that is m1 into m2 will be equal to 1 into minus 1 and that is equal to minus 1 so product is minus 1 so we say that the Diagnose are perpendicular that is they intersect at 90 degrees we know that Diagnose bisect each other and they intersect at 90 degrees so we say that Diagnose bisect at 90 degrees so the fourth condition is also proved so the four points satisfy all the four conditions thus they form a square which is the required answer this completes our session hope you enjoyed this session