Which is an equivalent expression
1 answer:
You might be interested in
Let's assign these coordinate points as P1(-3,2), P2(-2,-1), P3(2,-1) and P4(3,3). When you plot these points (shown in the picture), they are widely scattered. Therefore, they must not be a trend. Since you forgot to include the options, I would just make a smart guess that these coordinates form a shape which is quadrilateral. I just have to make use of what I have. From the given, I could possibly determine the area of the quadrilateral.
First, you need to plot the points in a cartesian plane as what is shown at the top of the attached picture. Then, you make a matrix wherein the x coordinates are placed in the first row, and the y coordinates in the second row (shown as the black box in the picture). Make sure that you arrange the points in order (whether clockwise or counterclockwise).
Once you create the matrix, you find its determinant. This is done by multiplying the numbers diagonally. The equation will be: arrow down - arrow up. The solution is as follows:
Determinant = [(-3×-1) + (-2×-1) + (2×3)] - [(2×-2) + (-1×2) + (-1×3)] = 20
Then, we use the equation for area of any polygon: A = 1/2*determinant
A = 1/2*20 = 10 square units
Thus, the output of the point plotted together creates a quadrilateral with an area of 10 square units.
First, you need to plot the points in a cartesian plane as what is shown at the top of the attached picture. Then, you make a matrix wherein the x coordinates are placed in the first row, and the y coordinates in the second row (shown as the black box in the picture). Make sure that you arrange the points in order (whether clockwise or counterclockwise).
Once you create the matrix, you find its determinant. This is done by multiplying the numbers diagonally. The equation will be: arrow down - arrow up. The solution is as follows:
Determinant = [(-3×-1) + (-2×-1) + (2×3)] - [(2×-2) + (-1×2) + (-1×3)] = 20
Then, we use the equation for area of any polygon: A = 1/2*determinant
A = 1/2*20 = 10 square units
Thus, the output of the point plotted together creates a quadrilateral with an area of 10 square units.