The vertices of quadrilateral MATH have coordinates M(-4,2), A(-1,-3), T(9,3) and H(6,8). Prove that quadrilateral MATH is a parallelogram. Prove that quadrilateral MATH is a rectangle. [The use of the set of axes below is optional]

The quadrilateral MATH is a rectangle because the opposite sides are equal and parallel (MH=AT=11.66 /MA=HT=5.83) and its internal angles are equal to 90°.

Quadrilaterals

There are different types of quadrilaterals, for example: square, rectangle, rhombus, trapezoid and parallelogram. Each type is defined accordingly to its length of sides and angles. For example, in a rectangle, the opposite sides are equal and parallel and their interior angles are equal to 90°.

Vector Magnitude Formula

It is possible to find the length of the sides from the vector magnitude |v|, since the magnitude of a vector represents the distance between two points.

[tex]|V|=\sqrt{(x_{2}-x_{1})^2+ \sqrt{(y_{2}-y_{1})^2[/tex]

The name of the rectangle is MATH, then its sides are: MA, AT, HT and MH.

First, you should find the length of these sides. Thus,

Find |MA|

[tex]|MA|=\sqrt{(-1-(-4))^2+(-3-2)^2} \\ \\ \|MA|=\sqrt{(-1+4)^2+(-5)^2}\\ \\|MA|=\ \sqrt{(3)^2+(-5)^2}\\ \\ |MA|=\sqrt{9+25} \\ \\ |MA|=\sqrt{34}=5.83[/tex]

Find |AT|

[tex]|AT|=\sqrt{(9-(-1))^2+(3-(-3))^2} \\ \\ \|AT|=\sqrt{(9+1)^2+(3+3)^2}\\ \\|AT|=\ \sqrt{(10)^2+(6)^2}\\ \\ |AT|=\sqrt{100+36} \\ \\ |AT|=\sqrt{136}=11.66[/tex]

Find |HT|

[tex]|HT|=\sqrt{(6-9))^2+(8-3)^2} \\ \\ \|HT|=\ \sqrt{(-3)^2+(-5)^2}\\ \\ |HT|=\sqrt{9+25} \\ \\ |HT|=\sqrt{34}=5.83[/tex]

Find |MH|

[tex]|MH|=\sqrt{(6-(-4))^2+(8-2)^2} \\ \\ |MH|=\sqrt{(6+4)^2+(6)^2} \\ \\ \|MH|=\ \sqrt{(10)^2+(6)^2}\\ \\ |MH|=\sqrt{100+36} \\ \\ |MH|=\sqrt{136}=11.66[/tex]

The previous calculations show that the opposite sides present the same lengths, since: MH=AT=11.66 and MA=HT=5.83. See the attached image.