Coordinates of a point, components of a vector and midpoint of a segment

Coordinates of a point on the plane

Let's see how vectors are used to assign coordinates to the points in the plane.

We consider a fixed point in the plane $O$ (known as origin), and a basis $B=\{\overrightarrow{u}, \overrightarrow{v}\}$ of $V_2$ (Space vector of dimension $2$).

Let's remember that a basis of $V_2$ are two linearly independent vectors. The set formed by $O$ and $B=\{\overrightarrow{u}, \overrightarrow{v}\}$ constitutes a reference system in the plane, since it allows us to determine the position of any other points on the plane.

This is because any other points $P$ on the plane determines along with point $O$ a vector $\overrightarrow{OP}$. Let $(p_1,p_2)$ be the components of the vector in basis $B$. Then $(p_1,p_2)$ are the coordinates of point $P$ in the reference system $R=\{O;\overrightarrow{u}, \overrightarrow{v}\}$ and we write $P =(p_1,p_2)$.

The procedure to find the coordinates of point $P$ in a given reference system is the following :

1. From the points $O$ and $P$ we determine the vector $\overrightarrow{OP}$

2. We express the vector $\overrightarrow{OP}$ as a linear combination of the vectors of the basis$B=\{\overrightarrow{u}, \overrightarrow{v}\}$, that is to say, $\overrightarrow{OP}=p_1 \cdot \overrightarrow{u}+p_2 \cdot \overrightarrow{v}$

3. $P=(p_1,p_2)$

Express point $P$ of the drawing in the reference system $R =\{O;\overrightarrow {u}, \overrightarrow{v}\}$.

• We draw the vector $\overrightarrow{OP}$:

• We express the vector $\overrightarrow{OP}$ as a linear combination of the vectors of the basis $B=\{\overrightarrow{u}, \overrightarrow{v}\}$:

• We obtain $\overrightarrow{OP}=\overrightarrow{u}+2\overrightarrow{v}$ and therefore the coordinates of the point $P$ are $P = (1 , 2)$

From now on we will consider, as reference system $R$, the one formed by the origin of coordinates $O = (0, 0)$ and the canonical basis of $V_2$ $B =\{\overrightarrow{i},\overrightarrow{j}\}$.

Components of a vector determined by two points

Let's see now the way to determine the components of a vector if we know the coordinates of its endpoints:

Let $P =(p_1,p_2)$ and $Q = (q_1,q_2)$ be two points of the plane, and $\overrightarrow{PQ}$ the vector that goes from $P$ to $Q$. Then the components of the vector $\overrightarrow{PQ}$ are $\overrightarrow{PQ}=(q_1-p_1,q_2-p_2)$.

Given $P = (2, 6)$ and $Q = (-3, 9)$. The components of the vector $\overrightarrow{PQ}$ are:$\overrightarrow{PQ}= (-3 - 2, 9 - 6) = (-5, 3)$

Applying a vector to a point

Given a point $P$ and a vector $\overrightarrow{v}$, the result of applying the vector to the point is a new point $Q$ placed in the direction of $\overrightarrow{v}$ and at a distance $|\overrightarrow{v}|$. (module of the vector $\overrightarrow{v}$)

The coordinates of this new point $Q$ are calculated from those of $P =(p_1,p_2)$ and $\overrightarrow{v}=(v_1,v_2)$ thus $$Q = P +\overrightarrow{v}=(p_1+v_1,p_2+v_2)$$

NOTE: It is very important to bear in mind that this addition operation only makes sense between a point and a vector. We must never add two points, and the result of adding two vectors is another vector and not a point!

Considering the following figure, determine the coordinates of point $P$ of the figure, the result of applying the vector $\overrightarrow{v}$ to the point $A$.

We begin by calculating the components of the vector $\overrightarrow{v}$:$$\overrightarrow{v} = (2 - (-1), 4-2) = (3, 2)$$ Since $P$ is the result of applying the vector $\overrightarrow{v}$ to the point $A$ we have,$$P=A+\overrightarrow{v}=(0,4)+(3,2)=(3,6)$$

Midpoint of a segment

Let's consider now a segment with endpoints $A = (a_1,a_2)$ and $B = (b_1,b_2)$. Let $M =(m_1,m_2)$ be the midpoint of the above mentioned segment. Obviously the above mentioned point satisfies that $\overrightarrow{AB}=2\cdot \overrightarrow{AM}$, or that $(b_1-a_1,b_2-a_2)=2\cdot (m_1-a_1,m_2-a_2)$

Separating component by component we obtain: $$\begin{array}{rcl} b_1-a_1 & = & 2 \cdot (m_1-a_1) \\ b_2-a_2 &=& 2\cdot (m_2-a_2) \end{array}$$ and isolating we have: $$\begin{array}{rcl} m_1 & = & \displaystyle \frac{a_1+b_1}{2}\\ m_2 &=& \displaystyle \frac{a_2+b_2}{2} \end{array}$$ So that we can calculate the coordinates of the midpoint of a segment from the coordinates of its endpoints.

Considering the points $A = (-3, 7)$ and $B = (1, 2)$ find the midpoint of the segment that they determine.

Applying the previous formulas we have: $$\begin{array}{rcl} m_1 & = & \displaystyle \frac{a_1+b_1}{2}= \frac{-3+2}{2}=-1\\ m_2 &=& \displaystyle \frac{a_2+b_2}{2}=\frac{7+2}{2}=\frac{9}{2} \end{array}$$ Therefore the midpoint of the segment $AB$ is $M = (-1, \displaystyle \frac{9}{2})$