Given the coordinates of points $A(-2, 1)$, $B(1, 0)$, and $C(2, 4)$, find the coordinates of point $E$ such that quadrilateral $ABEC$ is a parallelogram.

1. Problem Description

Given the coordinates of points $A(-2, 1)$, $B(1, 0)$, and $C(2, 4)$, find the coordinates of point $E$ such that quadrilateral $ABEC$ is a parallelogram.

2. Solution Steps

In a parallelogram, opposite sides are parallel and equal in length. In parallelogram $ABEC$, we have $AB \parallel CE$ and $AC \parallel BE$. Also, $\vec{AB} = \vec{CE}$ and $\vec{AC} = \vec{BE}$.

Let $E = (x, y)$. We can use $\vec{AB} = \vec{CE}$ to find the coordinates of $E$.

$\vec{AB} = B - A = (1 - (-2), 0 - 1) = (3, -1)$

$\vec{CE} = E - C = (x - 2, y - 4)$

Since $\vec{AB} = \vec{CE}$, we have:

$x - 2 = 3$ and $y - 4 = -1$

$x = 3 + 2 = 5$

$y = -1 + 4 = 3$

Therefore, $E = (5, 3)$.

We can also use $\vec{AC} = \vec{BE}$.

$\vec{AC} = C - A = (2 - (-2), 4 - 1) = (4, 3)$

$\vec{BE} = E - B = (x - 1, y - 0)$

Since $\vec{AC} = \vec{BE}$, we have:

$x - 1 = 4$ and $y - 0 = 3$

$x = 4 + 1 = 5$

$y = 3$

Therefore, $E = (5, 3)$.

3. Final Answer

The coordinates of point $E$ are $(5, 3)$.