Question

what is the most specific name for the figure? a(a, -b) b(-a, -b) c(-2a, b) d(0, b) square rectangle rhombus parallelogram

Explanation:

Step1: Use the slope formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\)

For side \(AB\): \(A(a,-b)\), \(B(-a,-b)\), \(m_{AB}=\frac{-b-(-b)}{-a - a}=0\)
For side \(BC\): \(B(-a,-b)\), \(C(-2a,b)\), \(m_{BC}=\frac{b-(-b)}{-2a-(-a)}=\frac{2b}{-a}=-\frac{2b}{a}\)
For side \(CD\): \(C(-2a,b)\), \(D(0,b)\), \(m_{CD}=\frac{b - b}{0-(-2a)} = 0\)
For side \(DA\): \(D(0,b)\), \(A(a,-b)\), \(m_{DA}=\frac{-b - b}{a-0}=-\frac{2b}{a}\)
Since \(m_{AB}=m_{CD}\) and \(m_{BC}=m_{DA}\), opposite sides are parallel.

Step2: Check if it is a rhombus, rectangle or square

For a rectangle, adjacent sides should be perpendicular (\(m_1\times m_2=- 1\)). \(m_{AB}\times m_{BC}=0\times(-\frac{2b}{a}) = 0
eq - 1\)
For a rhombus, all sides should be equal. Let's use the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
\(AB=\sqrt{(a + a)^2+(-b + b)^2}=2|a|\)
\(BC=\sqrt{(-2a + a)^2+(b + b)^2}=\sqrt{a^{2}+4b^{2}}\)
\(AB
eq BC\)
Since opposite sides are parallel, it is a parallelogram.

Answer:

Parallelogram