 Hi and welcome to the session, let us discuss the following question, question says, find the area of the quadrilateral whose vertices taken in order are minus 4, minus 2, minus 3, minus 5, 3, minus 2 and 2, 3. Let us now start the solution, let A minus 4, minus 2, B minus 3, minus 5, C 3, minus 2 and D 2, 3 are vertices of the quadrilateral. This is the quadrilateral given to us. Now by joining A to C, we will get two triangles, triangle ABC and triangle ADC. We know that area of the triangle found by the vertices x1, y1, x2, y2 and x3, y3 is given by 1 upon 2 multiplied by x1 multiplied by y2 minus y3 plus x2 multiplied by y3 minus y1 plus x3 multiplied by y1 minus y2. Now in this figure clearly we can see area of quadrilateral ABCD is equal to area of triangle ABC plus area of triangle ADC. So first of all we will find out area of these two triangles. Now first of all let us find out area of triangle ABC. Let us draw triangle ABC separately and assume that point A is represented by x1, y1, point B is represented by x2, y2 and point C is represented by x3, y3. So clearly we can see x1 is equal to minus 4, y1 is equal to minus 2, x2 is equal to minus 3, y2 is equal to minus 5, x3 is equal to 3 and y3 is equal to minus 2. Now we can find area of triangle ABC by using this formula. Now we get area of triangle ABC is equal to half multiplied by x1, x1 we know is equal to minus 4 multiplied by y2 minus y3, we know y2 is equal to minus 5 and y3 is equal to minus 2. So we can write minus 5 minus 2 here plus x2 multiplied by y3 minus y1 we know x2 is equal to minus 3, minus 3 multiplied by y3, y3 we know is equal to minus 2 minus y1, y1 is equal to minus 2. So we can write minus 2 minus minus 2 plus x3 we know x3 is equal to 3 multiplied by y1 minus y2, y1 is equal to minus 2 minus y2 is equal to minus 5. So we can write minus 2 minus minus 5. Now simplifying we get area of triangle ABC is equal to 1 upon 2 multiplied by minus 4 multiplied by minus 5 plus 2 plus minus 3 multiplied by minus 2 plus 2 plus 3 multiplied by minus 2 plus 5 here negative and negative sign will become positive here, here also negative and negative sign will become positive similarly here negative and negative sign will become positive. Now simplifying further we get 1 upon 2 multiplied by minus 4 multiplied by minus 3 we know minus 5 plus 2 is equal to minus 3, now this term will become 0 we know minus 2 plus 2 is equal to 0 and 0 multiplied by minus 3 is equal to 0. Now here we will get 3 multiplied by 3 which is equal to 9, now 1 upon 2 multiplied by 12 plus 9 we know minus 4 multiplied by minus 3 is equal to 12, now this is further equal to 21 upon 2 square units. So we get area of triangle ABC is equal to 21 upon 2 square units, now let us find out area of triangle ABC, let us assume that point x1 y1 represents vertex A, point x2 y2 represents vertex C and point x3 y3 represent vertex D, coordinates of A C and D are given to us, so clearly we can see value of x1 is equal to minus 4 in this case, y1 is equal to minus 2, x2 is equal to 3, y2 is equal to minus 2, x3 is equal to 2, y3 is equal to 3. So we can write area of triangle ADC is equal to 1 upon 2 multiplied by x1, x1 we know is equal to minus 4 multiplied by y2 minus y3, we know y2 is equal to minus 2 and y3 is equal to 3, so we can write minus 2 minus 3 plus x2 multiplied by y3 minus y1, we know x2 is equal to 3 and y3 is equal to 3, so 3 minus y1 is equal to minus 2, so 3 minus minus 2 plus x3, we know x3 is equal to 2 multiplied by y1 minus y2, y1 we know is equal to minus 2 and y2 is equal to minus 2, so we can write minus 2 minus minus 2, now simplifying we get 1 upon 2 multiplied by minus 4 multiplied by minus 5, we know minus 2 minus 3 is equal to minus 5 plus 3 multiplied by 3 plus 2 plus 3 multiplied by minus 2 plus 2, here 2 negative signs will change into positive sign and here also 2 negative signs will change into positive sign. Now we get 1 upon 2 multiplied by we know minus 4 multiplied by minus 5 is equal to 20, so 20 plus 3 plus 2 is 5 and 5 multiplied by 3 is equal to 15 plus we know minus 2 plus 2 is equal to 0 and 0 multiplied by 3 is 0, so finally we get 35 upon 2 square units area of triangle ADC is equal to 35 upon 2 square units, now we know area of quadrilateral ABCD is equal to area of triangle ABC plus area of triangle ADC, we know area of triangle ABC is equal to 21 upon 2 square units and area of triangle ADC is equal to 35 upon 2 square units, so adding these two values we will get area of quadrilateral ABCD, now this is equal to 56 upon 2 square units, now we will cancel common factor 2 from numerator and denominator both and get 28 square units, so required area of quadrilateral is equal to 28 square units, so 28 square units is our required answer this completes the session hope you understood the session take care and have a nice day.