After plotting the quadrilateral in a Cartesian plane, you can see that it is not a particular quadrilateral. Hence, you need to divide it into two triangles. Let's take ABC and ADC.
The area of a triangle with vertices known is given by the matrix
M =
Area = 1/2· | det(M) |
= 1/2· | x₁·y₂ - x₂·y₁ + x₂·y₃ - x₃·y₂ + x₃·y₁ - x₁·y₃ |
= 1/2· | x₁·(y₂ - y₃) + x₂·(y₃ - y₁) + x₃·(y₁ - y₂) |
Therefore, the area of ABC will be:
A(ABC) = 1/2· | (-5)·(-5 - (-6)) + (-4)·(-6 - 7) + (-1)·(7 - (-5)) |
= 1/2· | -5·(1) - 4·(-13) - 1·(12) |
= 1/2 | 35 |
= 35/2
Similarly, the area of ADC will be:
A(ABC) = 1/2· | (-5)·(5 - (-6)) + (4)·(-6 - 7) + (-1)·(7 - 5) |
= 1/2· | -5·(11) + 4·(-13) - 1·(2) |
= 1/2 | -109 |
<span> = 109/2</span>
The total area of the quadrilateral will be the sum of the areas of the two triangles:
A(ABCD) = A(ABC) + A(ADC)
= 35/2 + 109/2
= 72
Answer:
(-4,1)
Step-by-step explanation:
The above formulas do not hold for r = 1. For r = 1, the sum of n terms of the Geometric Progression is Sn
n
= na.
(ii)When the numerical value of r is less than 1 (i.e., - 1 < r < 1), then the formula Sn
n
= a(1−rn)(1−r)
(
1
−
r
n
)
(
1
−
r
)
is used.
(iii) When the numerical value of r is greater than 1 (i.e., r > 1 or, r < -1) then the formula Sn
n
= a(rn−1)(r−1)
(
r
n
−
1
)
(
r
−
1
)
is used.
(iv) When r = 1, then Sn
n
= a + a + a + a + a + .................... to n terms = na.
(v) If l is the last term of the Geometric Progression, then l = arn−1
n
−
1
.
Therefore, Sn
n
= a(1−rn1−r
1
−
r
n
1
−
r
) = (a−arn1−r
a
−
a
r
n
1
−
r
) = a−(arn−1)r(1−r)
a
−
(
a
r
n
−
1
)
r
(
1
−
r
)
= a−lr1−r
a
−
l
r
1
−
r
Thus, Sn
n
= a−lr1−r
a
−
l
r
1
−
r
Or, Sn
n
= lr−ar−1
l
r
−
a
r
−
1
, r ≠ 1.
You stop at 2 decimal points. If the 3rd decimal is bigger than 5, add 1 to the 2nd and if it's less than 5, keep it as it is.
For example:
2.663 --> 2.66
6.788 --> 6.79