Question

Show that points P(1,-2), Q(5,2), R(3,-1), S(-1,-5) are the vertices of a parallelogram

Answer 1

Formual used: 
1) Distance Between two points : $A(x_{1},y_{1}) \:\:and \:\: B(x_{2},y_{2})$
$\sqrt{ (x_{2}-x_{1})^{2} +(y_{2}-y_{1})^{2}}$

Steps:
1) 
[tex]PQ= \sqrt{(5-1)^{2}+(2-(-2))^{2}} = 4 \sqrt{2} \:\:units \\ \\ QR = \sqrt{(5-3)^{2}+(2-(-1))^{2}} = \sqrt{13} \:\:units \\ \\ RS= \sqrt{(3-(-1))^{2}+(-1-(-5))^{2}} = 4 \sqrt{2} \:\: units \\ \\ PS = \sqrt{(1-(-1))^{2}+(-2+5)^{2}} = \sqrt{13} \:\: units [/tex]

We observe that Opposite sides are equal .

2) Diagonals: 

$PR= \sqrt{(1-3)^{2} + (-2+1)^{2} } = \sqrt{5} \:units \\ \\ QS = \sqrt{(5+1)^{2}+ (2+5)^{2}} = \sqrt{85} units$

Since, Diagonals are un -equal and opposite sides are equal .
=> PQRS is a parallelogram . 

Answer 2

Answer:

pls mark as Brainlist answer