 Hello and welcome to the session. In this session we will discuss how can we use coordinates to prove simple geometric theorems algebraically. Now suppose we are given four coordinate points that is these four coordinate points. Now if we want to prove or disprove that the figure defined by these four given points is a parallelogram. Now we know that in a parallelogram the diagonals bisect each other and if in a coordinate the diagonals bisect each other then that parallelogram is a parallelogram. Now here we will prove that the figure defined by four given points is a parallelogram and is algebraically using coordinates. Let us plot these points of the coordinate plane. Now first point is having coordinates 3, 2. So this is the point with coordinates 3, 2. Then this is the point with coordinates 3 minus 1. Next this is the point with coordinates minus 1, minus 1 and this is the point with coordinates minus 1, 2. Now let us join all these points. Now let us name these points as a, b, c and d and by joining all these points we get a quadrilateral. Now a, c and b, d are the two diagonals. Now let us name the point of intersection as p. Now let us find the red points of these two diagonals algebraically. Now we know that red point of a line segment joining two 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. Using this midpoint formula we can find the midpoints of the two diagonals. So midpoint of diagonal a, c is given by the ordered pair 3 plus or minus 1 that is 3 minus 1 whole upon 2, 2 plus or minus 1 that is 2 minus 1 whole upon 2 that is equal to the ordered pair 2 upon 2, 1 upon 2 which is equal to the ordered pair 1, 1 upon 2. Similarly we can find which point of diagonal b, d which is given by the ordered pair 3 plus or minus 1 that is 3 minus 1 whole upon 2, minus 1 plus 2 whole upon 2 on solving. This is equal to ordered pair 1, 1 upon 2. Now from these calculations you can see that midpoint of diagonal a, c is same as midpoint of diagonal b, d. So midpoint is same that means the diagonals intersect each other. So a, c, d is a parallelogram as we have proved that the figure to find parallelogram 4 points is a parallelogram. Needle further proof of this proof that the given 4 points form a rectangle. For this we need to prove these three conditions. First is diagonals intersect each other. Second, length of diagonals is equal and third is length of opposite sides is equal. Now earlier we have proved condition 1 that is diagonals of this given quadrilateral intersect each other. Now we have to prove the second condition. So we will prove the second condition by using distance formula. Now according to distance formula the distance between any two points x1, y1 and x2, y2 is given by square root of x2 minus x1 whole square plus y2 minus y1 whole square. So length of diagonal a, c is given by square root of minus 1 minus 3 whole square plus minus 1 minus 2 whole square that is equal to square root of minus 4 whole square that is 16 plus minus 3 whole square that is 9 which is equal to square root of 25 that is equal to 5. So length of diagonal a, c, s, 5, e and x similarly we can find length of diagonal bd which will be equal to square root of minus 1 minus 3 whole square plus 2 minus of minus 1 that is 2 plus 1 whole square this is equal to square root of minus 1 minus 3 that is minus 4 whole square which is 16 plus 3 whole square that is 9 which is equal to square root of 25 that is again 5. So length of diagonal bd is also 5 units so length of diagonal ac is equal to length of diagonal bd is equal to 5 units as we have proved the second condition that is length of diagonal is equal. Now after proving condition 2 we will prove the third condition that is length of opposite sides is equal. Now let us find length of all sides using distance formula so length of side ab will be equal to square root of 3 minus 3 whole square plus minus 1 minus 2 whole square and on solving this is equal to 3. Now let us find length of side bc so this is equal to square root of minus 1 minus 3 whole square plus minus 1 minus of minus 1 that is plus 1 whole square which is equal to root 16 that is equal to 4. Now length of side cd is equal to square root of minus 1 minus of minus 1 that is plus 1 whole square plus 2 minus of minus 1 that is 2 plus 1 whole square and on solving this is equal to 3. Similarly length of side da is equal to 4 so from these calculations we can see that length of side ab is equal to length of side cd is equal to 3 units and length of side bc is equal to length of side da is equal to 4 units thus opposite sides are equal. So third condition is also satisfied so the four given points form a rectangle that is abcd is a rectangle. Now let us consider one more example proof of this proof point with coordinates 1 root 3 lies on the circle with center at origin and passing through the point 0 2 now here we are given that the circle passes through the point 0 2 at its center is origin that is its center is at the point 0 0 now we know that distance from the center to any point on the circle is equal to radius of the circle so from these three points we can find radius of the circle by using distance formula so radius r is equal to square root of x2 minus x1 whole square that is 0 minus 0 whole square plus y2 minus y1 whole square that is 0 minus 2 whole square that is equal to 0 square plus minus 2 whole square which is equal to square root of 4 that is equal to 2 so radius r is equal to 2 units now point with coordinates 1 root 3 will lie on the circle if its distance from the center is equal to the radius of the circle that is 2 so let us find the distance between the center with coordinates 0 0 at this point with coordinates 1 root 3 now let us denote center by point 0 and this particular point by the point p so distance op is equal to square root of 1 minus 0 whole square plus root 3 minus 0 whole square on solving op is equal to p is equal to 2 therefore op is equal to the radius of the circle r which is equal to 2 units thus point p with coordinates 1 root 3 lies on the circle now let us see an alternate method for proving this first of all we will find equation of the circle now we know that equation of circle with center at origin and radius r is given by x square plus y square is equal to r square now in the north example it was given that circle passes through the point 0 2 so this point will lie on the circle which means this point will satisfy this equation so putting x is equal to 0 and y is equal to 2 in this equation we have 0 square plus 2 square is equal to r square which implies 2 square is equal to r square this gives r is equal to 2 so taking positive square root we have obtained r is equal to 2 units so we have got the radius so we put r is equal to 2 in the given equation and we get this equation so equation of the circle with center at origin and which passes through the point 0 2 is given by x square plus y square is equal to 4 now we have to prove or disprove that this point with coordinates 1 root 3 lie on this circle now let this be equation 1 now if this point that is the point with coordinates 1 root 3 satisfies equation 1 then it lies on this circle otherwise not so we put x is equal to 1 and y is equal to root 3 in equation 1 and we get 1 square plus root 3 won't square is equal to 4 which implies 1 plus 3 is equal to 4 this implies 4 is equal to 4 which is true so the point with coordinates 1 root 3 satisfies equation 1 so this point lies on this circle given by equation 1 hence proved so in this session we have discussed how to use coordinates to prove simple geometric theorems as typically and this completes our session hope you all have enjoyed the session