Question 554928
We can calculate slopes of MA, AT, TH, HM, MT, and AH
For MA, slope = {{{(4-(-6))/(5-3)=10/2=5}}}
For AT, slope = {{{(-6-(-10))/(3-0)=4/3}}}
For TH, slope = {{{(0-(-10))/(2-0)=10/2=5}}}
For HM, slope = {{{(4-0)/(5-2)=4/3}}}
The fact that opposite sides of quadrilateral MATH have the same slope, means those pairs of opposite sides are parallel. That proves that it is a parallelogram.
If the adjacent sides were perpendicular, we would have four right angles, and it would be a rectangle (or maybe even that special kind of rectangle that we call square). If the sides were perpendicular, the product of their slopes would be -1.
However, {{{(5)(4/3)=20/3}}} is not -1. So, there are no right angles in MATH. Math is not a square or a rectangle.
Could it be a rhombus? If it were a rhombus, the diagonals would be perpendicular.
Let's calculate the slope of the diagonals
For MT, slope = {{{(4-(-10))/(5-0)=14/5}}}
For AH, slope = {{{(-6-0)/(3-2)=-6/1=-6}}}
The product of the slopes, {{{-6(14/5)=-84/5}}} , is not -1, so the diagonals are not perpendicular, and MATH is not a rhombus.
Quadrilateral MATH is a parallelogram. It is neither a square, nor a rectangle, nor a rhombus.