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Show that points P(1,-2),Q(5,2),R(3,-1),...

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

Text Solution

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`P(1,-2),Q(5,2),R(3,-1)` and `S(-1,-5)`
By distance formula
`PQ=sqrt((5-1)^(2)+[2-(-2)]^(2))`
`=sqrt(4^(2)+4^(2))`
`=sqrt(16+16)`
`sqrt(32)`
`=sqrt(2xx2xx2xx2xx2)`
`=4sqrt(2)`…………1
`QR=sqrt((3-5)^(2)+(-1-2)^(2))`
`=sqrt((-2)^(2)+(-3)^(2))`
`=sqrt(4+9)`
`=sqrt(13)` ...2
`RS=sqrt((-1-3)^(2)+[-5-(-1)]^(2)`
`=sqrt((-4)^(2)+(-4)^(2))`
`=sqrt(16+16)`
`=sqrt(32)`
`=sqrt(2xx2xx2xx2xx2)`
`=4sqrt(2)`....... 3
`PS=sqrt((-1s-1)^(2)+[-5-(-2)]^(2))`
`=sqrt((-2)^(2)+(-5+2)^(2))`
`=sqrt((-2)^(2)+(-3)^(2))`
`=sqrt(4+9)`
`sqrt(13)`...........4
In `squarePQRS`
`PQ=RS`....[From 1 and 3]
`QR=PS`........[From 2 and 4]
`square PQRS` is a prallelogram .......(A quadrilateral is a parallelogram if both the pairs of its opposite sides are congruent.)
`P(1,-2),Q(5,2),R(3,-1)` and `S(-1,-5)` are the vertices of a parallelogram.
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