This is a trapezoid because the line through (0,2) and (6,4) is parallel to the line through (-2, -3) and (10, 1)- they both have slope 1/3. In order that the figure we get moving one point still be a trapezoid, the point must stay on that line. The equation of a line through (6, 4) with slope 1/3 is y= (1/3)(x- 6)+ 4= (1/3)x+ 2. The new point, to which (0, 2) is moved, (x, y), must satisfy y= (1/3)x+ 2.
The distance from (6, 4) to (10, 1) is \(\displaystyle \sqrt{(6- 10)^2+ (4- 1)^2}= \sqrt{25}= 5\). In order that this new trapezoid be "isosceles" we must have the other "non-parallel" side, from (-2, -3) to (x, y), of the same length: \(\displaystyle \sqrt{(x+2)^2+ (y+ 3)^2}= 5\) or \(\displaystyle (x+ 2)^2+ (y+ 3)^2= 25\).
Solve \(\displaystyle y= (1/3)x+ 2\) and \(\displaystyle (x+ 2)^2+ (y+ 3)^2= 25\) for x and y.
(This doesn't really require any "Calculus".)
