show that point P (1, - 2), Q(5,2), R (3,-1), S(-1,-5) are the vertices of a parallelogram. ( Can anyone solve this sum with the distance formula?)
Answers
Answer:
used:
1) Distance Between two points : A(x_{1},y_{1}) \:\:and \:\: B(x_{2},y_{2})A(x
1
,y
1
)andB(x
2
,y
2
)
\sqrt{ (x_{2}-x_{1})^{2} +(y_{2}-y_{1})^{2}}
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
Steps:
1)
\begin{gathered}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\end{gathered}
PQ=
(5−1)
2
+(2−(−2))
2
=4
2
units
QR=
(5−3)
2
+(2−(−1))
2
=
13
units
RS=
(3−(−1))
2
+(−1−(−5))
2
=4
2
units
PS=
(1−(−1))
2
+(−2+5)
2
=
13
units
We observe that Opposite sides are equal .
2) Diagonals:
\begin{gathered}PR= \sqrt{(1-3)^{2} + (-2+1)^{2} } = \sqrt{5} \:units \\ \\ QS = \sqrt{(5+1)^{2}+ (2+5)^{2}} = \sqrt{85} units\end{gathered}
PR=
(1−3)
2
+(−2+1)
2
=
5
units
QS=
(5+1)
2
+(2+5)
2
=
85
units
Since, Diagonals are un -equal and opposite sides are equal .
=> PQRS is a parallelogram .
Step-by-step explanation:
i hope it helps you