Question

A parallelogram whose vertices have coordinates R(1, -1), S(6, 1), T(8, 5), and U(3, 3) has a shorter diagonal of ___ . 5 √(13) √(97)

Answer 1

Answer:

$|SU|=\sqrt{13}$

Step-by-step explanation:

The given parallelogram has vertices R(1, -1), S(6, 1), T(8, 5), and U(3, 3) .

Recall the distance formula;

We use the distance formula to determine the length of the diagonals.

For diagonal R(1,-1) and T(8,5), We have;

$|RT|=\sqrt{(8-1)^2+(5--1)^2}$

$|RT|=\sqrt{(7)^2+(6)^2}$

$|RT|=\sqrt{49+36}$

$|RT|=\sqrt{85}$

For the diagonal S(6,1) U(3,3)

$|SU|=\sqrt{(6-3)^2+(5-3)^2}$

$|SU|=\sqrt{(3)^2+(2)^2}$

$|SU|=\sqrt{9+4}$

$|SU|=\sqrt{13}$

Therefore the shorter diagonal is:

$|SU|=\sqrt{13}$