Question

if (-9, -3), (-6,3), and (5,3) are the coordinates of three vertices (corners) of a parallelogram, determine the coordinates of three different points that could serve as the fourth vertex. the fourth vertex could lie at the points . (type ordered pairs. use a comma to separate answers as needed. simplify your answers.)

Explanation:

Step1: Recall the property of parallelogram

In a parallelogram, the mid - points of the diagonals coincide. Let the points be \(A(-9,-3)\), \(B(-6,3)\), \(C(5,3)\). There are three possible cases for the fourth vertex \(D(x,y)\).

Case 1: Diagonal \(AC\) and \(BD\)

The mid - point of \(AC\) is \((\frac{-9 + 5}{2},\frac{-3+3}{2})=(-2,0)\). The mid - point of \(BD\) is \((\frac{-6 + x}{2},\frac{3 + y}{2})\). Then \(\frac{-6 + x}{2}=-2\) and \(\frac{3 + y}{2}=0\). Solving \(\frac{-6 + x}{2}=-2\) gives \(-6+x=-4\), so \(x = 2\). Solving \(\frac{3 + y}{2}=0\) gives \(y=-3\).

Case 2: Diagonal \(AB\) and \(CD\)

The mid - point of \(AB\) is \((\frac{-9-6}{2},\frac{-3 + 3}{2})=(-\frac{15}{2},0)\). The mid - point of \(CD\) is \((\frac{5 + x}{2},\frac{3 + y}{2})\). Then \(\frac{5 + x}{2}=-\frac{15}{2}\) and \(\frac{3 + y}{2}=0\). Solving \(\frac{5 + x}{2}=-\frac{15}{2}\) gives \(5+x=-15\), so \(x=-20\). Solving \(\frac{3 + y}{2}=0\) gives \(y = - 3\).

Case 3: Diagonal \(AD\) and \(BC\)

The mid - point of \(BC\) is \((\frac{-6 + 5}{2},\frac{3+3}{2})=(-\frac{1}{2},3)\). The mid - point of \(AD\) is \((\frac{-9 + x}{2},\frac{-3 + y}{2})\). Then \(\frac{-9 + x}{2}=-\frac{1}{2}\) and \(\frac{-3 + y}{2}=3\). Solving \(\frac{-9 + x}{2}=-\frac{1}{2}\) gives \(-9+x=-1\), so \(x = 8\). Solving \(\frac{-3 + y}{2}=3\) gives \(-3+y = 6\), so \(y=9\).

Answer:

\((2,-3),(-20,-3),(8,9)\)