Question

question 4
1 pts
quadrilateral math has coordinates
m(1,1), a(-2,5), t(3,5), and h(6,1).
what is the most precise name for the quadrilateral?
square
parallelogram
rectangle
rhombus

Explanation:

Step1: Calculate the slopes of the sides

The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\).
For \(MA\) with \(M(1,1)\) and \(A(-2,5)\):
\(m_{MA}=\frac{5 - 1}{-2 - 1}=\frac{4}{-3}=-\frac{4}{3}\)
For \(AT\) with \(A(-2,5)\) and \(T(3,5)\):
\(m_{AT}=\frac{5 - 5}{3-(-2)} = 0\)
For \(TH\) with \(T(3,5)\) and \(H(6,1)\):
\(m_{TH}=\frac{1 - 5}{6 - 3}=\frac{-4}{3}\)
For \(HM\) with \(H(6,1)\) and \(M(1,1)\):
\(m_{HM}=\frac{1 - 1}{1 - 6}=0\)

Step2: Check for parallel sides

Since \(m_{MA}=m_{TH}=-\frac{4}{3}\) and \(m_{AT}=m_{HM}=0\), opposite sides are parallel.

Step3: Check for right - angles (not necessary for parallelogram, but for rectangle, square, rhombus)

The product of slopes of adjacent sides:
\(m_{MA}\times m_{AT}=-\frac{4}{3}\times0 = 0\) (but we also check side lengths)
The length formula is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
\(MA=\sqrt{(-2 - 1)^2+(5 - 1)^2}=\sqrt{9 + 16}=\sqrt{25} = 5\)
\(AT=\sqrt{(3-(-2))^2+(5 - 5)^2}=\sqrt{25+0}=5\)
\(TH=\sqrt{(6 - 3)^2+(1 - 5)^2}=\sqrt{9 + 16}=\sqrt{25}=5\)
\(HM=\sqrt{(1 - 6)^2+(1 - 1)^2}=\sqrt{25+0}=5\)
But for a square, adjacent sides should be perpendicular (\(m_1\times m_2=- 1\)). Here \(m_{MA}\times m_{AT}=0
eq - 1\). For a rhombus, adjacent sides are not perpendicular (but we already know it's a parallelogram with all sides equal). For a rectangle, adjacent sides should be perpendicular. Since \(m_{MA}\times m_{AT}=0\) (not \(-1\)), it's not a rectangle or square.

Answer:

parallelogram