Math, asked by rainirodriquez, 6 months ago

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

Answered by rudrakshasahu123
1

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

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