Question

If P(-2, 4), Q(4,8), R(10,5) and S(4, 1) are
vertices of quadrilateral show that it is a
parallelogram.​

Answer 1

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Answer 2

Yes; P(-2, 4), Q(4,8), R(10,5) and S(4, 1) are the vertices of a parallelogram

Step-by-step explanation:

Given:  P(-2, 4), Q(4,8), R(10,5) and S(4, 1) are vertices of a quadrilateral

To prove: PQRS is a parallelogram

As we know the distance formula

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ where $(x_1,y_1),(x_2,y_2)$ are the end points coordinated

therefore the distance of   P(-2, 4), Q(4,8)

$PQ= \sqrt{(4-(-2))^2+(8-4)^2} = \sqrt{6^2+4^2} =\sqrt{36+16} =\sqrt{52}$

similarly the distance of Q(4,8) , R(10,5)  

$QR= \sqrt{(10-4)^2+(5-8)^2} = \sqrt{6^2+(-3)^2} =\sqrt{36+9} =\sqrt{45}$

similarly the distance of R(10,5) , S(4,1)

$RS= \sqrt{(4-10)^2+(1-5)^2} = \sqrt{(-6)^2+(-4)^2} =\sqrt{36+16} =\sqrt{52}$

and the distance of P (-2,4) , S(4,1)

$PS= \sqrt{(4-(-2))^2+(1-4)^2} = \sqrt{6^2+(-3)^2} =\sqrt{36+9} =\sqrt{45}$

as we can see PQ=RS and QR=PS

as the opposite sides of a parallelogram are equal then it is  a parallelogram

Hence PQRS is a parallelogram

#Learn more

If P(-5,-3), Q(-4,-6), R(2,-3) and S(1,2) are the vertices of a quadrilateral PQRS, then find its area.

brainly.in/question/2266418