Question

15 three vertices of a quadrilateral are given: a(2,3), b(7,5), and c(9,1). where would point d be located to create a parallelogram? input your answer as a point.

Explanation:

Step1: Recall property of parallelogram

In a parallelogram, the mid - points of the diagonals coincide. Let the diagonals be $AC$ and $BD$ or $AD$ and $BC$ or $AB$ and $CD$. 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 $A(2,3)$, $B(7,5)$ and $C(9,1)$. If the diagonals are $AC$ and $BD$, and the mid - point of $AC$ is the same as the mid - point of $BD$.
The mid - point of $AC$ is $(\frac{2 + 9}{2},\frac{3+1}{2})=(\frac{11}{2},2)$.
Let $D(x,y)$, the mid - point of $BD$ is $(\frac{7 + x}{2},\frac{5 + y}{2})$.

Step2: Set up equations

We have the following system of equations based on the mid - point equality:
$\frac{7 + x}{2}=\frac{11}{2}$ and $\frac{5 + y}{2}=2$.
From $\frac{7 + x}{2}=\frac{11}{2}$, we get $7+x = 11$, so $x=11 - 7=4$.
From $\frac{5 + y}{2}=2$, we get $5 + y=4$, so $y=4 - 5=-1$.

Answer:

$(4,-1)$