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No. For example, a 12x1 and a 4x3 quadrilateral both have an area of 12, but they are not congruent.
12x1=12 12x2=24 12x3=36 12x4=48 12x5=60 12x6=72
Ok.The midpoint formula: [(x1 + x2)/2, (y1 + y2)/2]So for instance if your coordinates were endpoint : (-8,10) and the Midpoint: (-2,6)By substituting the given How_do_you_find_an_endpoint_of_a_line_if_you_are_given_an_endpoint_and_the_midpointinto the formula we have:(x1 + -8)/2 = -2 and (y1 + 10)/2 = 6x1 - 8 = -4 and y1 + 10 = 12x1 -8 + 8 = -4 + 8 and y1 + 10 - 10 = 12 - 10x1 = 4 and y1 = 2so the endpoints coordinates are ( 4, 2)
Ok.The midpoint formula: [(x1 + x2)/2, (y1 + y2)/2]So for instance if your coordinates were endpoint : (-8,10) and the Midpoint: (-2,6)By substituting the given values into the formula we have:(x1 + -8)/2 = -2 and (y1 + 10)/2 = 6x1 - 8 = -4 and y1 + 10 = 12x1 -8 + 8 = -4 + 8 and y1 + 10 - 10 = 12 - 10x1 = 4 and y1 = 2so the endpoints coordinates are ( 4, 2)