# Determining Coordinates

In **2D coordinate geometry**, each point corresponds to a pair of numbers \((a,b)\) such that the first number \(a\) gives the \(x\text{-coordinate}\), the second number \(b\) gives the \(y\text{-coordinate}\) of the point. We use these coordinates to determine the placement of the point as follows:

#### Contents

## Steps for Determining Coordinates

Given a point in the coordinate plane, how do we determine the \(x\text{-coordinate}\) and \(y\text{-coordinate}\) of the point? It may be useful to follow these steps to help solve the problem:

- Draw a picture
- Label the picture
- Mark the point or other desired quantity the problem is asking for
- Draw any additional lines or points that may help you find the solution

Given point \((a,b)\), let's consider what happens when we reflect the point in the following ways.

- Reflect the point about the \(x\)-axis: this gives the point \((a,-b)\) with the same \(x\)-coordinate and with \(y\)-coordinate multiplied by \(-1\).

- Reflect the point about the \(y\)-axis: this gives the point \((-a,b)\) with the same \(y\)-coordinate and with \(x\)-coordinate multiplied by \(-1\).

- Reflect the point about the origin \((0, 0)\): this gives the point \((-a,-b)\) with both \(x\) and \(y\)-coordinates multiplied by \(-1\).

- Reflect the point about the line \(y=x\): this gives the point \((b,a)\), with \(x\) and \(y\)-coordinates swapped.

- Reflect the point about the line \(y=-x\): this gives the point \((-b,-a)\), with \(x\) and \(y\) coordinates swapped and multiplied by \(-1\).

Let's illustrate these concepts of reflection in the following example.

## Consider the point \(A=(2, 5).\) What are the coordinates of the reflections of \(A\) about

- the \(x\)-axis
- the \(y\)-axis
- the origin \((0, 0)\)
- the line \(y=x\)
- the line \(y=-x,\)
respectively?

Solution:The reflections of a point \((2, 5)\) about

- the \(x\)-axis is \((2, -5)\)
- the \(y\)-axis is \((-2, 5)\)
- the origin \((0, 0)\) is \((-2, -5)\)
- the line \(y=x\) is \((5, 2)\)
- the line \(y=-x,\) is \((-5, -2).\) \( _\square \)

## Determining coordinates for translations

A set of points undergoes a **translation** for each point is mapped from \((x,y)\) to \((x+a, y+b)\) for some fixed numbers \(a\) and \(b\). To determine the coordinates of any point after this translation, add \(a\) to the \(x\)-coordinate and \(b\) to the \(y\)-coordinate of the point.

## A parallel translation \((x, y) \longrightarrow (x-3, y+b)\) moves the point \((-2, 5)\) to the point \((a, 7).\) What is \(a+b?\)

Solution:The parallel translation \((x, y) \longrightarrow (x-3, y+b)\) moves the point \((-2, 5)\) to \[(-2-3, 5+b).\] Equating this with the point \((a, 7)\) given in the problem, we have \[-2-3=a, 5+b=7,\] which implies \(a+b=-5+2=-3.\) \( _\square \)

## Determining coordinates in geometric figures

Suppose that we are given a geometric figure and would like to determine the coordinates of one or more of the points of the figure. In this case, the following additional steps are useful:

- remember geometric properties of the figure in the problem
- draw a diagram for the problem that illustrate these properties

## Parallelogram \(ABCD\) has vertices \[A=(-2, a), B=(b, 0), C=(3, 1), D=(1, 3).\] What is \(a+b?\)

Solution:Since \(ABCD\) is a parallelogram, it must be true that the midpoints of \(\overline{AC}\) and \(\overline{BD}\) are the same. In other words, \[\left(\frac{-2+3}{2}, \frac{a+1}{2}\right)=\left(\frac{b+1}{2}, \frac{0+3}{2}\right).\] This implies \(-2+3=b+1\) or \(b=0,\) and \(a+1=0+3\) or \(a=2.\)Therefore, \(a+b=2+0=2.\) \( _\square \)

## Parallelogram \(ABCD\) has vertex \(A=(0, 2).\) If the midpoints of sides \(\overline{AB}\) and \(\overline{BC}\) are \(X=(-1, 0)\) and \(Y=(2, -1),\) respectively, what are the coordinates of vertex \(D?\)

Solution:Let \(B=(a, b),\) then \[X=\left(\frac{0+a}{2}, \frac{2+b}{2}\right)=(-1, 0) \Rightarrow a=-2, b=-2.\] Similarly, let \(C=(c, d),\) then \[Y=\left(\frac{-2+c}{2}, \frac{-2+d}{2}\right)=(2, -1) \Rightarrow c=6, d=0.\]Thus, we have \(B=(-2, -2)\) and \(C=(6, 0).\) Now, let \(D=(e, f),\) then since the midpoints of of \(\overline{AC}\) and \(\overline{BD}\) are the same in parallelogram \(ABCD,\) \[\left(\frac{0+6}{2}, \frac{2+0}{2}\right)=\left(\frac{-2+e}{2}, \frac{-2+f}{2}\right) \Rightarrow e=8, f=4.\]

Therefore, \(D=(8, 4).\) \( _\square \)

**Cite as:**Determining Coordinates.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/determining-coordinates/