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The quadrilateral PQRS appears to be a parallelogram. (In fact, in the snapshot above, it shows up to it is in a rhombus, yet if we draw a square ABCD v two consecutive sides much much shorter than the various other two sides, we deserve to see that the political parties of PQRS need not be equal.)To prove that PQRS is a parallelogram, us will examine that side PQ is parallel to SR and that QR is parallel to PS.
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v the works with of A, B, C, and D as given above, we haveegineqnarray*P & = & left( fracx_1 + x_22, fracy_1 + y_22 ight) \Q & = & left( fracx_2 + x_32, fracy_2 + y_32 ight) \R & = & left( fracx_3 + x_42, fracy_3 + y_42 ight) \S & = & left( fracx_4 + x_12, fracy_4 + y_12 ight) \endeqnarray*So the slope of the heat PQ isegineqnarray*fracDelta yDelta x & = & fracfracy_2 + y_32 - fracy_1 + y_22fracx_2 + x_32 - fracx_1 + x_22 \ & = & frac(y_2 + y_3) - (y_1 + y_2)(x_2 + x_3) - (x_1 + x_2) \ & = & fracy_3 - y_1x_3 - x_1endeqnarray*The steep of the line SR isegineqnarray*fracDelta yDelta x & = & fracfracy_3 + y_42 - fracy_4 + y_12fracx_3 + x_42 - fracx_4 + x_12 \ & = & frac(y_3 + y_4) - (y_4 + y_1)(x_3 + x_4) - (x_4 + x_1) \ & = & fracy_3 - y_1x_3 - x_1endeqnarray*These slopes space the same; therefore, the present PQ and SR room parallel. We deserve to compute the slopes of present QR and PS to uncover that both slopes are equal to$$ fracy_4 - y_2x_4 - x_2. $$Therefore, QR and also PS room parallel together well. Due to the fact that the square PQRS has actually two bag of parallel sides, that is a parallelogram.

You are watching: Which statement proves that pqrs is a parallelogram?

Solution:Alternative equipment to (b)

We might simplify the algebraic work of this proof slightly by applying a change to the quadrilateral ABCD before we begin. Suppose that, by using a translation and a rotation together necessary, we relocate the quadrilateral ABCD so that the vertex A is in ~ the beginning $(0, 0)$, and also the vertex B is what on the x-axis, in ~ the suggest $(x_2, 0)$. We define $(x_3, y_3)$ and also $(x_4, y_4)$ to be the new coordinates of the vertices C and also D, respectively.

We now haveegineqnarray*P & = & left( fracx_22, 0 ight) \Q & = & left( fracx_2 + x_32, fracy_32 ight) \R & = & left( fracx_3 + x_42, fracy_3 + y_42 ight) \S & = & left( fracx_42, fracy_42 ight) \endeqnarray*So the steep of the heat PQ isegineqnarray*fracDelta yDelta x & = & fracfracy_32 - 0fracx_2 + x_32 - fracx_22 \ & = & fracy_3 - 0(x_2 + x_3) - x_2 \ & = & fracy_3x_3endeqnarray*The steep of the heat SR isegineqnarray*fracDelta yDelta x & = & fracfracy_3 + y_42 - fracy_42fracx_3 + x_42 - fracx_42 \ & = & frac(y_3 + y_4) - y_4(x_3 + x_4) - x_4 \ & = & fracy_3x_3endeqnarray*These slopes space the same, therefore the currently PQ and SR space parallel. We can compute the slopes of present QR and PS to find that both slopes are equal to$$ fracy_4x_4 - x_2. $$So QR and PS space parallel as well. Because the quadrilateral PQRS has two pairs of parallel sides, that is a parallelogram.


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Typeset might 4, 2016 in ~ 18:58:52.

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