Show that it is a point reflection at A

mathmari

Well-known member
MHB Site Helper
Hey!!

We have the points $Z=(-1,1)$, $A=(-1,6)$ and $B=(3,4)$.

Let $\delta$ be the rotation with center $Z$ and $\delta (A)=B$.

Let $C$ be the point on the circumcircle of the triangle $ABZ$ such that the segment $\overline{CZ}$ goes through the center of circumcircle.

Let $\gamma$ be a rotation with center $C$ and $\gamma (B)=A$.

Show that $\gamma\circ\delta$ is a point reflection at $A$.

To show that do we have to show that there is exactly one fixed point? Or do we have to determine the matrices of the transformation and consider th determinant of the matrix?

Klaas van Aarsen

MHB Seeker
Staff member
Hey mathmari !!

The composition of 2 rotations is a rotation, so it will always have exactly one fixed point (unless we get identity or a translation).
So we can't show it like that.
Furthermore a rotation always has a matrix with determinant +1, so we cannot show it with the determinant either.

What is special about a point reflection is that we can write it as the matrix $-I$ combined with a translation.
So we should e.g. verify what the resulting matrix is.
Or alternatively we can find the center of rotation and verify that a point is transformed to the opposite side of that center.
Or alternatively we can find the images of 2 different points and verify that the corresponding line segments intersect each other in the middle.

mathmari

Well-known member
MHB Site Helper
What is special about a point reflection is that we can write it as the matrix $-I$ combined with a translation.
So we should e.g. verify what the resulting matrix is.
Or alternatively we can find the center of rotation and verify that a point is transformed to the opposite side of that center.
Or alternatively we can find the images of 2 different points and verify that the corresponding line segments intersect each other in the middle.
We have that $$\delta \begin{pmatrix}x \\ y\end{pmatrix}= \begin{pmatrix}3/5 & 4/5 \\ -4/5 & 3/5\end{pmatrix}\left [\begin{pmatrix}x \\ y\end{pmatrix} -\begin{pmatrix}-1 \\ 1\end{pmatrix} \right ]+\begin{pmatrix}-1 \\ 1\end{pmatrix}$$ and $$\gamma \begin{pmatrix}x \\ y\end{pmatrix}= \begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}\left [\begin{pmatrix}x \\ y\end{pmatrix} -\begin{pmatrix}1 \\ 5\end{pmatrix} \right ]+\begin{pmatrix}1 \\ 5\end{pmatrix}$$ Then the composition $\gamma\circ\delta$ is given by \begin{align*}\left (\gamma \circ\delta \right ) \begin{pmatrix}x \\ y\end{pmatrix}&=\gamma \left ( \begin{pmatrix}3/5 & 4/5 \\ -4/5 & 3/5\end{pmatrix}\left [\begin{pmatrix}x \\ y\end{pmatrix} -\begin{pmatrix}-1 \\ 1\end{pmatrix} \right ]+\begin{pmatrix}-1 \\ 1\end{pmatrix}\right )\\ & =\gamma \left ( \begin{pmatrix}3/5 & 4/5 \\ -4/5 & 3/5\end{pmatrix}\begin{pmatrix}x+1 \\ y-1\end{pmatrix} +\begin{pmatrix}-1 \\ 1\end{pmatrix}\right )\\ & =\gamma \begin{pmatrix}\frac{3x}{5}+\frac{4y}{5}-\frac{6}{5}\\ -\frac{4x}{5}+\frac{3y}{5}-\frac{2}{5}\end{pmatrix}\\ & =\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}\left [\begin{pmatrix}\frac{3x}{5}+\frac{4y}{5}-\frac{6}{5}\\ -\frac{4x}{5}+\frac{3y}{5}-\frac{2}{5}\end{pmatrix} -\begin{pmatrix}1 \\ 5\end{pmatrix} \right ]+\begin{pmatrix}1 \\ 5\end{pmatrix} \\ & =\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix}\frac{3x}{5}+\frac{4y}{5}-\frac{6}{5}-1\\ -\frac{4x}{5}+\frac{3y}{5}-\frac{2}{5}-2\end{pmatrix} +\begin{pmatrix}1 \\ 5\end{pmatrix}\\ & =\begin{pmatrix}-\frac{3x}{5}-\frac{4y}{5}+\frac{11}{5}\\ \frac{4x}{5}-\frac{3y}{5}+\frac{12}{5}\end{pmatrix} +\begin{pmatrix}1 \\ 5\end{pmatrix}\\ & =\begin{pmatrix}-\frac{3x}{5}-\frac{4y}{5}+\frac{11}{5}+1\\ \frac{4x}{5}-\frac{3y}{5}+\frac{12}{5}+5\end{pmatrix}\\ & =\begin{pmatrix}-\frac{3x}{5}-\frac{4y}{5}+\frac{16}{5}\\ \frac{4x}{5}-\frac{3y}{5}+\frac{37}{5}\end{pmatrix} \end{align*} Is that correct so far? How could we continue?

Klaas van Aarsen

MHB Seeker
Staff member
The matrix for $\delta$ is correct.

But how did you get that $\gamma$ has matrix $-I$?
Since $\gamma(B)=A$ and since $C$ is not halfway between $A$ and $B$ that cannot be correct.
It is not a point reflection. Instead we need to show that $(\gamma\circ\delta)$ is a point reflection.

mathmari

Well-known member
MHB Site Helper
But how did you get that $\gamma$ has matrix $-I$?
Since $\gamma(B)=A$ and since $C$ is not halfway between $A$ and $B$ that cannot be correct.
It is not a point reflection. Instead we need to show that $(\gamma\circ\delta)$ is a point reflection.
Is the point $C=\begin{pmatrix}1 \\ 5\end{pmatrix}$ correct?

Klaas van Aarsen

MHB Seeker
Staff member
Is the point $C=\begin{pmatrix}1 \\ 5\end{pmatrix}$ correct?
That point is in between $A$ and $B$.
But $C$ is on the circumcircle of triangle $ABZ$, which implies that it cannot be between $A$ and $B$.

Klaas van Aarsen

MHB Seeker
Staff member
I've drawn a picture now:
\begin{tikzpicture}
\usetikzlibrary {angles,calc,quotes,intersections}

\draw[help lines] (-4,0) grid (4,7);
\draw[-latex] (-4,0) -- (4,0);
\draw[-latex] (0,0) -- (0,7);
\draw foreach \i in {-1,3} { (\i,0) node[below] {$\i$} };
\draw foreach \i in {1,4,6} { (0,\i) node[ left ] {$\i$} };

\coordinate[label=above:A] (A) at (-1,6);
\coordinate[label=right:B] (B) at (3,4);
\coordinate[label=below:Z] (Z) at (-1,1);
\coordinate (M) at (0.25,3.5);
\draw[name path=ZM, help lines] (Z) -- ($(Z)!6cm!(M)$);
\draw[name path=cc, help lines] (M) circle (2.8);
\path[name intersections={of = ZM and cc}] coordinate[label=right:C] (C) at (intersection-1);
\draw[thick] (A) -- (B) -- (Z) -- cycle;
\fill (M) circle (.04);
\draw pic [draw, shorten <=2pt, latex-, angle radius=2.5cm] {angle = A--C--B};
\draw pic [draw, shorten <=2pt, latex-, angle radius=5cm] {angle = B--Z--A};
\end{tikzpicture}

We can see that $C$ is not at $(1,5)$ can't we?

mathmari

Well-known member
MHB Site Helper
We can see that $C$ is not at $(1,5)$ can't we?
So, we have to find the equation of the circle that passes through the points $A,B,Z$.
Let the circle be $x^2+y^2+Dx+Ey+F=0$.
We substitute the three points andsolve the system.
\begin{align*}& Z : \ 1+1-D+E+F=0 \Rightarrow 2-D+E+F=0 \\ &A: \ 1+36-D+6E+F=0 \Rightarrow 37-D+6E+F=0 \\ &B : \ 9+16+3D+4E+F=0 \Rightarrow 25+3D+4E+F=0\end{align*}
Solving that system we get $D=-\frac{1}{2}, \ E=-7, \ F=\frac{9}{2}$.

So the equation of the circle is $x^2+y^2-\frac{1}{2}x-7y+\frac{9}{2}=0$.
Then we have to find the center of the circle,say $K$, and then we find the line that passes though $Z$ and $K$.
The intersection point of the circle and the line os the point $C$.

Is that correct?

Staff member
Correct yes.

mathmari

Well-known member
MHB Site Helper
Correct yes.
The equation of the circle in other form is $$\left (x-\frac{1}{4}\right )^2+\left (y-\frac{7}{2}\right )^2=\frac{125}{16}$$ So the center is $K\left (\frac{1}{4},\ \frac{7}{2}\right )$.
The line that passes through $K$ and $Z$ is $$y-1=\frac{ \frac{7}{2}-1}{\frac{1}{4}+1 }(x+1) \Rightarrow y-1=2(x+1)$$
The intersection point of the circle and the line (different from $Z$) is $C\left (\frac{3}{2}, \ 6\right )$.

Then we have that $$\gamma \begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}a& b \\ c & d\end{pmatrix}\left [\begin{pmatrix}x \\ y\end{pmatrix}-\begin{pmatrix}\frac{3}{2} \\ 6\end{pmatrix}\right ]+\begin{pmatrix}\frac{3}{2} \\ 6\end{pmatrix}$$ and we find the elements of the matrix using that $\gamma (B)=A$, right?

\begin{align*}\gamma (B)=A &\Rightarrow \begin{pmatrix}a& b \\ c & d\end{pmatrix}\left [\begin{pmatrix}3 \\ 4\end{pmatrix}-\begin{pmatrix}\frac{3}{2} \\ 6\end{pmatrix}\right ]+\begin{pmatrix}\frac{3}{2} \\ 6\end{pmatrix}=\begin{pmatrix}-1 \\ 6\end{pmatrix} \\ & \Rightarrow \begin{pmatrix}a& b \\ c & d\end{pmatrix}\begin{pmatrix}\frac{3}{2} \\ -2\end{pmatrix}+\begin{pmatrix}\frac{3}{2} \\ 6\end{pmatrix}=\begin{pmatrix}-1 \\ 6\end{pmatrix}\\ & \Rightarrow \begin{pmatrix}a& b \\ c & d\end{pmatrix}\begin{pmatrix}\frac{3}{2} \\ -2\end{pmatrix}=\begin{pmatrix}-\frac{5}{2} \\ 0\end{pmatrix}\\ & \Rightarrow \begin{pmatrix}\frac{3a}{2} -2b\\ \frac{3c}{2} -2d\end{pmatrix}=\begin{pmatrix}-\frac{5}{2} \\ 0\end{pmatrix}\end{align*}
From that we get $b=\frac{3a}{4}+\frac{5}{4}$ and $d=\frac{3c}{4}$

Do we give arbitrarily values to $a$ and $c$ to get a specific matrix?

If for example we take $a=c=1$ we get
So we get \begin{equation*}\gamma \begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}1& 2\\ 1 & \frac{3}{4}\end{pmatrix}\left [\begin{pmatrix}x \\ y\end{pmatrix}-\begin{pmatrix}\frac{3}{2} \\ 6\end{pmatrix}\right ]+\begin{pmatrix}\frac{3}{2} \\ 6\end{pmatrix}\end{equation*} Is that correct?

Klaas van Aarsen

MHB Seeker
Staff member
Correct except we do not give arbitrary values to $a$ and $c$.
Instead the matrix must be a rotation matrix.

mathmari

Well-known member
MHB Site Helper
Correct except we do not give arbitrary values to $a$ and $c$.
Instead the matrix must be a rotation matrix.
Ah it must be an orthogonal matrix, i.e. $M^TM=I$.
So we get \begin{equation*} \begin{pmatrix}a& \frac{3a}{4}+\frac{5}{4} \\ c & \frac{3c}{4}\end{pmatrix}^Τ\begin{pmatrix}a& \frac{3a}{4}+\frac{5}{4} \\ c & \frac{3c}{4}\end{pmatrix}=\begin{pmatrix}1& 0 \\ 0 & 1\end{pmatrix} \Rightarrow \begin{pmatrix}a& c\\ \frac{3a}{4}+\frac{5}{4} & \frac{3c}{4}\end{pmatrix}\begin{pmatrix}a& \frac{3a}{4}+\frac{5}{4} \\ c & \frac{3c}{4}\end{pmatrix}=\begin{pmatrix}1& 0 \\ 0 & 1\end{pmatrix}\Rightarrow \begin{pmatrix}a^2+c^2& \frac{3a^2}{4}+\frac{5a}{4} +\frac{3c^2}{4}\\ \frac{3a^2}{4}+\frac{5a}{4} +\frac{3c^2}{4} & \frac{25}{16} + \frac{15 a}{8} + \frac{9 a^2}{16} + \frac{9 c^2}{16}\end{pmatrix}=\begin{pmatrix}1& 0 \\ 0 & 1\end{pmatrix}\Rightarrow a=-\frac{3}{5}, \ c=\pm \frac{4}{5}\end{equation*}
Which value of $c$ do we have to get?

Klaas van Aarsen

MHB Seeker
Staff member
A 2x2 orthogonal matrix is either a rotation matrix or a reflection matrix.

mathmari

Well-known member
MHB Site Helper
A 2x2 orthogonal matrix is either a rotation matrix or a reflection matrix.
Ahh the determinant must be $+1$, right? So we have to use $c=-\frac{4}{5}$

So we have the matrix $\begin{pmatrix}-\frac{3}{5}& \frac{4}{5} \\ - \frac{4}{5} &- \frac{3}{5}\end{pmatrix}$, i.e \begin{equation*}\gamma \begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}-\frac{3}{5}& \frac{4}{5} \\ -\frac{4}{5} & - \frac{3}{5}\end{pmatrix}\left [\begin{pmatrix}x \\ y\end{pmatrix}-\begin{pmatrix}\frac{3}{2} \\ 6\end{pmatrix}\right ]+\begin{pmatrix}\frac{3}{2} \\ 6\end{pmatrix}\end{equation*}

Is everything correct so far?

Then do we have to find the matrix of $\gamma\circ\delta$ and show that it is an orthogonal matrix with determinant $-1$ ?

Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
There is still $\pm$ in the matrix.

If the matrix of $\gamma\circ\delta$ is orthogonal with determinant $-1$, then it is the matrix of a reflection in a line, which is not possible because the composition of 2 rotation matrices is a rotation matrix (or identity).

Instead we have to verify that $\gamma\circ\delta$ is a "point reflection", which is actually a rotation with angle $180^\circ$.

mathmari

Well-known member
MHB Site Helper
Instead we have to verify that $\gamma\circ\delta$ is a "point reflection", which is actually a rotation with angle $180^\circ$.
So we have to show that the determiannt of $\gamma\circ\delta$ is $+1$ ? What else so we have to show t get the desired result?

Klaas van Aarsen

MHB Seeker
Staff member
So we have to show that the determiannt of $\gamma\circ\delta$ is $+1$ ? What else so we have to show t get the desired result?
If the matrix has determinant $+1$, then it is a rotation and it could be any rotation.
However, it has to be a rotation by $180^\circ$, so that is not good enough.
What is the matrix of a rotation by $180^\circ$?

mathmari

Well-known member
MHB Site Helper
If the matrix has determinant $+1$, then it is a rotation and it could be any rotation.
However, it has to be a rotation by $180^\circ$, so that is not good enough.
What is the matrix of a rotation by $180^\circ$?
Ahh the matrix has to be $-I$.

We have the following :
\begin{align*}\left (\gamma \circ\delta \right ) \begin{pmatrix}x \\ y\end{pmatrix}&=\gamma \left ( \begin{pmatrix}3/5 & 4/5 \\ -4/5 & 3/5\end{pmatrix}\left [\begin{pmatrix}x \\ y\end{pmatrix} -\begin{pmatrix}-1 \\ 1\end{pmatrix} \right ]+\begin{pmatrix}-1 \\ 1\end{pmatrix}\right )\\ & =\gamma \left ( \begin{pmatrix}3/5 & 4/5 \\ -4/5 & 3/5\end{pmatrix}\begin{pmatrix}x+1 \\ y-1\end{pmatrix} +\begin{pmatrix}-1 \\ 1\end{pmatrix}\right )\\ & =\gamma \begin{pmatrix}\frac{3x}{5}+\frac{4y}{5}-\frac{6}{5}\\ -\frac{4x}{5}+\frac{3y}{5}-\frac{2}{5}\end{pmatrix}\\ & =\begin{pmatrix}-\frac{3}{5}& \frac{4}{5} \\ - \frac{4}{5} & - \frac{3}{5}\end{pmatrix}\left [\begin{pmatrix}\frac{3x}{5}+\frac{4y}{5}-\frac{6}{5}\\ -\frac{4x}{5}+\frac{3y}{5}-\frac{2}{5}\end{pmatrix}-\begin{pmatrix}\frac{3}{2} \\ 6\end{pmatrix}\right ]+\begin{pmatrix}\frac{3}{2} \\ 6\end{pmatrix}\\ & =\begin{pmatrix}-\frac{3}{5}& \frac{4}{5} \\ - \frac{4}{5} & - \frac{3}{5}\end{pmatrix}\begin{pmatrix}\frac{3x}{5}+\frac{4y}{5}-\frac{6}{5}-\frac{3}{2}\\ -\frac{4x}{5}+\frac{3y}{5}-\frac{2}{5}-6\end{pmatrix}+\begin{pmatrix}\frac{3}{2} \\ 6\end{pmatrix}\\ & =\begin{pmatrix}-\frac{3}{5}& \frac{4}{5} \\ - \frac{4}{5} & - \frac{3}{5}\end{pmatrix}\begin{pmatrix}\frac{3x}{5}+\frac{4y}{5}-\frac{27}{10}\\ -\frac{4x}{5}+\frac{3y}{5}-\frac{32}{5}\end{pmatrix}+\begin{pmatrix}\frac{3}{2} \\ 6\end{pmatrix}\\ & =\begin{pmatrix}-x-\frac{7}{2} \\ -y+6\end{pmatrix}+\begin{pmatrix}\frac{3}{2} \\ 6\end{pmatrix}\\ & =\begin{pmatrix}-x-2 \\ -y+12\end{pmatrix}\\ & =\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}+\begin{pmatrix}-2 \\ 12\end{pmatrix}\end{align*}
So the matrix is $-I$, that means that $\gamma \circ\delta$ is a rotation by $180^\circ$.

Is that correct and complete?

Klaas van Aarsen

MHB Seeker
Staff member
Correct.

The problem statement is: "Show that γ∘δ is a point reflection at A."
You have shown that it is a point reflection.
Is it at A?

mathmari

Well-known member
MHB Site Helper
The problem statement is: "Show that γ∘δ is a point reflection at A."
You have shown that it is a point reflection.
Is it at A?
So do we have to show that $(\gamma \circ\delta)(A)=A$ ?

Klaas van Aarsen

MHB Seeker
Staff member
So do we have to show that $(\gamma \circ\delta)(A)=A$ ?
Yes.

Alternatively we could have written the transformation as $(\gamma \circ\delta)(x) = -I (x-A)+A$.