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