Quadrilateral KATE has vertices K A T E (1,5), (4,7), (7,3), and (1, 1). a) Prove that KATE is a trapezoid. b) Prove that KATE is not an isosceles trapezoid.

Question

There is also supposed to be work on the graph. 

Answer 1

 

 

(a) The slope of KA is (7-5)/(4-1)=⅔, and the slope of TE is (3-(-1))/(7-1)=4/6=⅔, and this common slope makes the lines parallel. Also KE is vertical while the slope of AT is (3-7)/(7-4)=-4/3. So KE is not parallel to AT, and KATE qualifies as a trapezoid.

(b) KE has length 5-(-1)=6 (difference between y coordinates) and the x-coordinates are the same. The length of AT is √((7-4)²+(3-7)²)=√25=5, which is not 6, so KE is not the same length as AT, and the trapezoid is not isosceles.