Can anyone answer this using coordinate geometry:the diagonals of a rhombus are perpendicular?

Question

This is the hint provided :(Hint:Let the vertices be (0,0), (a,0), (a+b,c), and (b,c) Show that c^2=a^2-b^2.) plz help test tommorow and plz show work

Answer 1

The slope of the diagonal from 0,0 to a+b,c is c/a+b The other diagonal has the slope of c/b-a. Since they are perpendicular, c/a+b= -(a+b)/c When you cross multiply and foil, you should get c^2=a^2-b^2

Answer 2

I don't know if you have to all that. Let the rhombus corners be 0,a,b,(b+c). Draw a-b and
0-b+c, the short and long diagionals. They intersect at p. The proof consists of showing that the triangles opb=b+c,p,b and triangle oap=b+c,p,a. are congruent. Then, since triangle o,b,b+c is isoceles, angle opb=angle b+c,p,b and they are right angles. The rest follows swiftly.

Answer 3

Find slopes

(0,0) and (b, c); slope = c / b

(a,0) and (a+b,c); slope = c/b