Exercise set 2 part a and b please! **Shoelace Theorem:**Suppose the polygon \( P \) has vertices \( (a_1, b_1), (a_2, b_2), \ldots, (a_n, b_n) \) listed in clockwise order. Define \( a_{n+1} = a_1, b_{n+1} = b_1 \). Then, the area of \( P \) can be calculated using the formula:\[\text{area of } P = \frac{1}{2} \left| \sum_{k=1}^{n} \det \begin{bmatrix} a_k & a_{k+1} \\ b_k & b_{k+1} \end{bmatrix} \right|\]Note that this is the absolute value of the sum of determinants.---**Exercise Set 2:**Plot the points, draw the polygon, and calculate its area using the Shoelace Theorem.(a) \((1, 3), (2, 1), (5, 0), (6, 4), (4, 2)\)(b) \((0, 0), (4, -1), (7, 2), (6, 5), (3, 7), (1, 4)\)

Shoelace Theorem: Let us suppose the polygon P has vertices a1,b1,a2,b2, . . . . an,bn listed in clockwise order. Define an+1=a1, bn+1=b1. Then, the area of the polygon P can be calculated by using the formula given by: Area of P=12∑k=1ndetakak+1bkbk+1(a) The given points are 1, 3, 2, 1, 5, 0, 6, 4, 4, 2. Let us consider the points as, a1,b1=1, 3, a2,b2=2, 1, a3,b3=5, 0, a4,b4=6, 4, a5,b5=4, 2 The points are plotted in the graph shown below. The required polygon by joining the given points is shown below. (The points must be joined in clockwise order.)Now, the points in clockwise order are: a1,b1=1, 3, a2,b2=6, 4, a3,b3=5, 0, a4,b4=4, 2, a5,b5=2, 1 Then, the formula to calculate the area of the polygon P by using the Shoelace Theorem is given by: Area of P=12∑k=1ndetakak+1bkbk+1 Since 5 points are given, therefore, n=5. Then, we have: Area of P=12∑k=15detakak+1bkbk+1 where, an+1=a1 ⇒a5+1=a1 ⇒a6=a1bn+1=b1 ⇒b5+1=b1 ⇒b6=b1Then, we get: Area of P=12deta1a1+1b1b1+1+deta2a2+1b2b2+1+deta3a3+1b3b3+1+deta4a4+1b4b4+1+deta5a5+1b5b5+1 Area of P=12deta1a2b1b2+deta2a3b2b3+deta3a4b3b4+deta4a5b4b5+deta5a6b5b6 Area of P=12det1634+det6540+det5402+det4221+det2113 Area of P=124-18+0-20+10-0+4-4+6-1 Area of P=12-14+-20+10+0+5 Area of P=12-14-20+10+0+5 Area of P=12-19 Area of P=12×19 ∵ -a=a Area of P=192 sq. units Hence, the area is 192 sq. units.(b) The given points are 0, 0, 4, -1, 7, 2, 6, 5, 3, 7, 1, 4. Let us consider the points as, a1,b1=0, 0, a2,b2=4, -1, a3,b3=7, 2, a4,b4=6, 5, a5,b5=3, 7, a6,b6=1, 4 The points are plotted in the graph shown below. The required polygon by joining the given points is shown below. (The points must be joined in clockwise order.)Now, the points in clockwise order are: a1,b1=0, 0, a2,b2=1, 4, a3,b3=3, 7, a4,b4=6, 5, a5,b5=7, 2, a6,b6=4, -1 Then, the formula to calculate the area of the polygon P by using the Shoelace Theorem is given by: Area of P=12∑k=1ndetakak+1bkbk+1 Since 6 points are given, therefore, n=6. Then, we have: Area of P=12∑k=16detakak+1bkbk+1 where, an+1=a1 ⇒a6+1=a1 ⇒a7=a1bn+1=b1 ⇒b6+1=b1 ⇒b7=b1Then, we get: Area of P=12deta1a1+1b1b1+1+deta2a2+1b2b2+1+deta3a3+1b3b3+1+deta4a4+1b4b4+1+deta5a5+1b5b5+1+deta6a6+1b6b6+1 Area of P=12deta1a2b1b2+deta2a3b2b3+deta3a4b3b4+deta4a5b4b5+deta5a6b5b6+deta6a7b6b7 Area of P=12det0104+det1347+det3675+det6752+det742-1+det4010 Area of P=120-0+7-12+15-42+12-35+-7-8+0-0 Area of P=120+-5+-27+-23+-15+0 Area of P=120-5-27-23-15+0 Area of P=12-70 Area of P=12×70 ∵ -a=a Area of P=35 sq. units Hence, the area is 35 sq. units.