 Hello friends, let's discuss the following question. It says, draw a quadrilateral in Cartesian plane whose vertices are minus 4, 5, 0, 7, 5, minus 5 and minus 4, minus 2. Also find its area. Let us first understand the basic approach to solve this question. Divide the quadrilateral into two triangles and the area of triangle having vertices x1, y1, x2, y2 and x3, y3 is given by 1 by 2 into mod of x1 into y2 minus y3 plus x2 into y3 minus y1 plus x3 into y1 minus y2. So this will be the key idea. Let us now proceed on with the solution. The vertices are a minus 4, 5, b, 0, 7, 5, minus 5 and d minus 4, minus 2. Now we have to draw a quadrilateral having these vertices. So let us now draw the quadrilateral having these vertices. So we first plot these points on the Cartesian plane. Point A is minus 4, 5, x coordinate is minus 4 and y coordinate is 5. So as this point, then we have to plot the point b which is 0, 7, x coordinate is 0, y coordinate is 7. So it is this point. Then we have to plot the point 5 minus 5, x is 5, y is minus 5. So it is this coordinate. And then we have to plot the point b which is minus 4, minus 2. x coordinate is minus 4, y coordinate is minus 2. So it is this point. Now we join these points to get the quadrilateral. So this is the quadrilateral obtained by joining these vertices. Now we have to find the area of this quadrilateral for that we divide this quadrilateral into two triangles by joining the points b and d to make the diagonal bd. Now we will find the area of the triangle a, b, d and b, c, d by this formula and then we will add the two areas to get the area of the quadrilateral. Now the vertices of triangle a, b, d are minus 4, 5, 0, 7 and minus 4, minus 2. Now we find the area of the triangle a, b, d. The area of triangle a, b, d is equal to 1 by 2 into mod of x1 here x1 is minus 4 into y2 minus y3 where y2 is 7, y3 is minus 2 plus x2 where x2 is 0 into y3 minus y1 where y3 is minus 2 and y1 is 5. So it is minus 2 minus 5 plus x3 which is minus 4 into y1 minus y2 which is 5 minus 7 and this is equal to 1 by 2 into mod of minus 36 plus 8 that is 1 by 2 into 28 which is equal to 14. Let's call this as 1. Now we find the area of the triangle b, c, d, b, c, d. Now the vertices of triangle b, c, d are 0, 7, 5 minus 5 and minus 4 minus 2. Now the area of triangle b, c, d is equal to 1 by 2 into mod of x1 into y2 minus y3 here x1 is 0, y2 is minus 5, y3 is minus 2 so it is minus of minus 2 plus x2 here x2 is 5 into y3 minus y1 that is minus 2 minus 7 plus x3 here x3 is minus 4 into y1 minus y2. Here y1 is 7, y2 is minus 5 so it is 7 minus minus 5. Simplifying this we have this is equal to 1 by 2 into mod of minus 45 minus 48 and this is equal to 1 by 2 into 93 and this is equal to 46.5. Let's call this as 2. Now the area of quadrilateral a, b, c, d is equal to area of triangle a, b, d plus area of triangle b, c, d. Now the area of triangle a, b, d is 14 units and area of triangle b, c, d is 46.5 units square units and this is equal to 60.5. Since it's area the unit is square units hence area of quadrilaterally a, b, c, d is equal to 60.5 square units which is same as 121 upon 2 square units. This completes the question. Hope you enjoyed this session. Goodbye and take care.