Question

drag the ordered pair to the correct location on the graph.
the figure on the graph below is being used in a coordinate proof of the statement that the diagonals of a rhombus are perpendicular to one another.
what are the coordinates for vertex l in the figure?
$(2b,0)$ $(4a,0)$ $(4b,0)$ $(a + b,0)$ $(2a + 2b,0)$ $(2a,0)$

Explanation:

Step1: Recall the property of a rhombus

In a rhombus, the diagonals bisect each other.

Step2: Use the mid - point formula

The mid - point formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\). Let \(J(0,0)\) and \(L(x,0)\) (since \(L\) is on the \(x -\)axis, \(y = 0\)), and \(K(a,b)\). For the diagonal \(JK\) and \(JL\), the mid - point of \(JK\) is \((\frac{0 + a}{2},\frac{0 + b}{2})\) and the mid - point of \(JL\) is \((\frac{0+x}{2},0)\). Since the diagonals of a rhombus bisect each other, if we consider the symmetry of the rhombus about the mid - point of the diagonals. Another way: In a rhombus \(JKLM\), \(JK=KL\). Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), \(JK=\sqrt{(a - 0)^2+(b - 0)^2}=\sqrt{a^{2}+b^{2}}\). Also, since \(J(0,0)\) and \(L(x,0)\) and \(K(a,b)\), and by the property of rhombus (all sides are equal \(JK = KL\)). But a more straightforward approach is using the fact that in a rhombus, if we assume the center of the rhombus (mid - point of the diagonals). The center of the rhombus (mid - point of diagonal \(JL\) and \(KM\)): Let the mid - point of \(JL\) be \((\frac{0 + x}{2},0)\) and for a rhombus, if we consider the symmetry of the vertex \(K(a,b)\) with respect to the center of the rhombus. Since the diagonals bisect each other, if the center of the rhombus (mid - point of \(JL\)) is also the mid - point of \(KM\). But a simpler geometric approach: In a rhombus, if we assume the vector or the symmetry about the center. If we know that in a rhombus \(JKLM\), \(J(0,0)\) and \(K(a,b)\), then using the property that the diagonals bisect each other. Let the mid - point of \(JL\) be \(O\). The mid - point of \(JK\) (if we consider the wrong approach) is not. The correct approach is: In a rhombus, the diagonals bisect each other. Let \(JL\) be one diagonal and \(KM\) be the other diagonal. The mid - point of \(JL\) is \((\frac{0 + x}{2},0)\) and since the rhombus is symmetric about the mid - point of its diagonals. If we use the fact that in a rhombus \(JKLM\), \(JK\parallel LM\) and \(JL\) is a diagonal. Using the property of parallelograms (a rhombus is a parallelogram), for a parallelogram \(JKLM\) with \(J(0,0)\) and \(K(a,b)\), the mid - point of \(JL\) (where \(L=(2a,0)\)) gives that the mid - point of \(JL\) is \((a,0)\) which can also be related to the mid - point of \(KM\) (by the property of parallelograms: diagonals bisect each other).

Answer:

\((2a,0)\)