The formula for perimeter is p = 2 l + 2 w. If the perimeter p of a rectangle is 200 mm (millimeters) and the width w = 30 mm, w
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Answer:
Step-by-step explanation:
a). Given equation is y = 3x - 4
Table for the input-output values is,
x -1 0 1 2 3
y -7 -4 -1 2 5
Now we can plot these points on the graph (graph attached).
b). Equation is y = -2x + 3
Table for the input-output values will be,
x -1 0 1 2 3
y 5 3 1 -1 -3
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 =![\left[\begin{array}{ccc} x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20x_%7B1%7D%26y_%7B1%7D%261%5C%5Cx_%7B2%7D%26y_%7B2%7D%261%5C%5Cx_%7B3%7D%26y_%7B3%7D%261%5Cend%7Barray%7D%5Cright%5D%20)
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
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