Question

the vertices of a rectangle are (-2,-3), (3,-3), (3,6), and (-2,6). when the rectangle is graphed in the standard (x,y) coordinate plane, what fraction of the total area of the rectangle lies in quadrant iii? 1/12 1/9 2/15

Explanation:

Step1: Calculate the length and width of the rectangle

The length of the rectangle is the distance between \(x\) - coordinates of \((-2,-3)\) and \((3,-3)\). Using the formula \(d=\vert x_2 - x_1\vert\), we have \(l=\vert3-(-2)\vert = 5\).
The width of the rectangle is the distance between \(y\) - coordinates of \((3,-3)\) and \((3,6)\). Using the formula \(d=\vert y_2 - y_1\vert\), we have \(w=\vert6 - (-3)\vert=9\).
The area of the rectangle \(A = l\times w=5\times9 = 45\).

Step2: Calculate the area of the part in Quadrant III

The part of the rectangle in Quadrant III has a length (in \(x\) - direction) from \(x=-2\) to \(x = 0\), so \(l_{III}=\vert0-(-2)\vert = 2\). The width (in \(y\) - direction) from \(y=-3\) to \(y = 0\), so \(w_{III}=\vert0-(-3)\vert=3\).
The area of the part in Quadrant III \(A_{III}=l_{III}\times w_{III}=2\times3 = 6\).

Step3: Calculate the fraction

The fraction \(f=\frac{A_{III}}{A}\). Substitute \(A_{III} = 6\) and \(A = 45\) into the formula, we get \(f=\frac{6}{45}=\frac{2}{15}\).

Answer:

\(\frac{2}{15}\)