Question

three vertices of a parallelogram are shown in the figure below. give the coordinates of the fourth vertex.

Explanation:

Step1: Recall property of parallelogram

In a parallelogram, the mid - points of the diagonals coincide. Let the vertices of the parallelogram be \(A(-1,-6)\), \(B(-7,-3)\), \(C(-3,0)\) and the fourth vertex be \(D(x,y)\). There are three possible cases depending on which pairs of points form the diagonals.
Case 1: If the diagonals are \(AC\) and \(BD\).
The mid - point formula for two points \((x_1,y_1)\) and \((x_2,y_2)\) is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\).
The mid - point of \(AC\) is \((\frac{-1+( - 3)}{2},\frac{-6 + 0}{2})=(\frac{-4}{2},\frac{-6}{2})=(-2,-3)\).
The mid - point of \(BD\) is \((\frac{-7+x}{2},\frac{-3 + y}{2})\).
So, \(\frac{-7+x}{2}=-2\) and \(\frac{-3 + y}{2}=-3\).
Solving \(\frac{-7+x}{2}=-2\) gives \(-7+x=-4\), then \(x = 3\).
Solving \(\frac{-3 + y}{2}=-3\) gives \(-3 + y=-6\), then \(y=-3\).
Case 2: If the diagonals are \(AB\) and \(CD\).
The mid - point of \(AB\) is \((\frac{-1+( - 7)}{2},\frac{-6+( - 3)}{2})=(\frac{-8}{2},\frac{-9}{2})=(-4,-4.5)\).
The mid - point of \(CD\) is \((\frac{-3+x}{2},\frac{0 + y}{2})\).
So, \(\frac{-3+x}{2}=-4\) and \(\frac{y}{2}=-4.5\).
Solving \(\frac{-3+x}{2}=-4\) gives \(-3+x=-8\), then \(x=-5\).
Solving \(\frac{y}{2}=-4.5\) gives \(y=-9\).
Case 3: If the diagonals are \(AD\) and \(BC\).
The mid - point of \(BC\) is \((\frac{-7+( - 3)}{2},\frac{-3+0}{2})=(\frac{-10}{2},\frac{-3}{2})=(-5,-1.5)\).
The mid - point of \(AD\) is \((\frac{-1+x}{2},\frac{-6 + y}{2})\).
So, \(\frac{-1+x}{2}=-5\) and \(\frac{-6 + y}{2}=-1.5\).
Solving \(\frac{-1+x}{2}=-5\) gives \(-1+x=-10\), then \(x=-9\).
Solving \(\frac{-6 + y}{2}=-1.5\) gives \(-6 + y=-3\), then \(y = 3\).

Let's assume the most common case where the diagonals are \(AC\) and \(BD\).

Answer:

\((3,-3)\)