Question

A Parallelogram whose vértices have coordinates r(1,-1) s(6,1) t (8,5) u 3,3 have a shorter diagonal

Answer 1

Answer:

The two shortest diagonals seems to be RU and TS. A further explanation is provided below.

Step-by-step explanation:

The given coordinated are:

r = (1, 1)

s = (6, 1)

t = (8, 5)

u = (3, 3)

Now,

The distance between RU will be:

= $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

On putting the values, we get

= $\sqrt{(3-1)^2+(3+1)^2}$

= $\sqrt{(2)^2+(4)^2}$

= $\sqrt{4+16}$

= $\sqrt{20}$