Determining Coordinates

In 2D coordinate geometry, each point corresponds to a pair of numbers \((a,b)\) such that the first number \(a\) gives the \(x\)-coordinate and the second number \(b\) gives the \(y\)-coordinate of the point.

We use these coordinates to determine the placement of the point, as shown below:

Given a point in the coordinate plane, how do we determine the \(x\)-coordinate and \(y\)-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?

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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 \((-5, -2).\) \( _\square \)

A set of points undergoes a translation if 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 \((x, y) \rightarrow (x-3, y+b)\) moves the point \((-2, 5)\) to the point \((a, 7).\) What is \(a+b?\)

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The parallel translation \((x, y) \rightarrow (x-3, y+b)\) moves the point \((-2, 5)\) to

\[(-2-3, 5+b).\]

Equating this with the point \((a, 7)\) given in the problem, we have

\[-2-3=a,\quad 5+b=7,\]

which implies \(a+b=-5+2=-3.\) \( _\square \)

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 illustrates these properties.

Parallelogram \(ABCD\) has vertices

\[A=(-2, a),\ B=(b, 0),\ C=(3, 1),\ D=(1, 3).\]

What is \(a+b?\)

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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 \(D?\)

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Let \(B=(a, b),\) then

\[X=\left(\frac{0+a}{2}, \frac{2+b}{2}\right)=(-1, 0) \implies a=-2, b=-2.\]

Similarly, let \(C=(c, d),\) then

\[Y=\left(\frac{-2+c}{2}, \frac{-2+d}{2}\right)=(2, -1) \implies c=6, d=0.\]

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 \(ABCD,\)

\[\left(\frac{0+6}{2}, \frac{2+0}{2}\right)=\left(\frac{-2+e}{2}, \frac{-2+f}{2}\right) \implies e=8, f=4.\]

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