Answer:
Tell me if i am wrong. :)
Step-by-step explanation:
To solve for y:
Q.1 2x + y = -1
y = -1 - 2x
Q. 2. 4x - 5y = 7
y = 7 - 4x over -5
Check the picture below.
now, we're making an assumption that, the two blue shaded region are equal in shape, and thus if that's so, that area above the 14 is 6 and below it is also 6, 14 + 6 + 6 = 26.
so hmm if we simply get the area of the trapezoid and subtract the area of the yellow triangle and the area of the cyan triangle, what's leftover is what we didn't subtract, namely the shaded region.
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h~~=height\\ a,b=\stackrel{parallel~sides}{bases~\hfill }\\[-0.5em] \hrulefill\\ h=15\\ a=14\\ b=26 \end{cases}\implies A=\cfrac{15(14+26)}{2}\implies A=300 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Areas}}{\stackrel{trapezoid}{300}~~ - ~~\stackrel{yellow~triangle}{\cfrac{1}{2}(26)(9)}~~ - ~~\stackrel{cyan~triangle}{\cfrac{1}{2}(15)(6)}} \\\\\\ 300~~ - ~~117~~ - ~~45\implies 138\qquad \textit{blue shaded area}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%20%5Cbegin%7Bcases%7D%20h~~%3Dheight%5C%5C%20a%2Cb%3D%5Cstackrel%7Bparallel~sides%7D%7Bbases~%5Chfill%20%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20h%3D15%5C%5C%20a%3D14%5C%5C%20b%3D26%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B15%2814%2B26%29%7D%7B2%7D%5Cimplies%20A%3D300%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Areas%7D%7D%7B%5Cstackrel%7Btrapezoid%7D%7B300%7D~~%20-%20~~%5Cstackrel%7Byellow~triangle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%2826%29%289%29%7D~~%20-%20~~%5Cstackrel%7Bcyan~triangle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%2815%29%286%29%7D%7D%20%5C%5C%5C%5C%5C%5C%20300~~%20-%20~~117~~%20-%20~~45%5Cimplies%20138%5Cqquad%20%5Ctextit%7Bblue%20shaded%20area%7D)
When you arrange the N points in sequence around the polygon (clockwise or counterclockwise), the area is half the magnitude of the sum of the determinants of the points taken pairwise. The N determinants will also include the one involving the last point and the first one.
For example, consider the vertices of a triangle: (1,1), (2,3), (3,-1). Its area can be computed as
(1/2)*|(1*3-1*2) +(2*-1-3*3) +(3*1-(-1)*1)|
= (1/2)*|1 -11 +4| = 3
Answer:
Your question is in what base please?