Transformations of straight lines and other graphs
Learn how to find transformations of straight lines and other graphs for Year 11 Specialist Mathematics in Queensland (QCAA). A two-by-two matrix moves every point of a graph to an image point, and this topic shows how a line, circle or curve is carried to its image.
You will learn to map two points, use the inverse-substitution method for an image equation, apply dilations, reflections and rotations, and use the determinant to see when a line stays a line — the groundwork for composing transformations later in the course.
Theory
A matrix transformation sends every point \((x,y)\) of a graph to an image point \((x',y')=A\begin{pmatrix}x\\y\end{pmatrix}\). In Year 11 Specialist Mathematics (QCAA, Queensland) a straight line always maps to another straight line, and this page shows how to find the image equation — by mapping two points or by the inverse substitution method.
A linear transformation of the plane is given by a \(2\times2\) matrix \(A\). It sends a point \((x,y)\) to its image \((x',y')\) by \(\begin{pmatrix}x'\\y'\end{pmatrix}=A\begin{pmatrix}x\\y\end{pmatrix}\). Every linear transformation fixes the origin.
The basic transformations each have a standard matrix: a dilation \((x,y)\to(ax,by)\) is \(\begin{pmatrix}a&0\\0&b\end{pmatrix}\); a reflection in \(y=x\) is \(\begin{pmatrix}0&1\\1&0\end{pmatrix}\) and in the \(y\)-axis is \(\begin{pmatrix}-1&0\\0&1\end{pmatrix}\); a rotation of \(90^\circ\) anticlockwise about \(O\) is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\).
To find the image of a line there are two reliable methods. Map two points: transform any two points on the line and join the images. Or use inverse substitution: because \((x,y)=A^{-1}(x',y')\), replace \(x\) and \(y\) in the equation by the inverse map and simplify. A non-singular \(A\) always sends a line to a line.
The determinant controls the effect. If \(\det A\neq0\) the image of a line is a line (and areas scale by \(|\det A|\)). If \(\det A=0\) the matrix is singular and the whole plane collapses onto one line through the origin. The same substitution method sends a circle to an ellipse.
The image of a point under the matrix \(A\):
The standard origin-fixed matrices:
The inverse substitution that gives the image equation of a curve:
How to find the image of a line
- Choose your method: for a quick answer map two points on the line; for the image equation directly, use inverse substitution.
- Two-point method: pick two convenient points (often the intercepts), multiply each by \(A\), then find the gradient and equation of the line through the two images.
- Inverse-substitution method: write \((x,y)=A^{-1}(x',y')\), substitute these into the original equation, and simplify to a relation in \(x'\) and \(y'\).
- State the image: drop the dashes and give the image equation; check a known point (such as the mapped intercept) satisfies it.
Choose two points on the line and map each one. Take the \(y\)-intercept \((0,-1)\):
| \(P'\) | \(=\) | \(\begin{pmatrix}2&0\\0&1\end{pmatrix}\begin{pmatrix}0\\-1\end{pmatrix}\) |
| \(=\) | \(\begin{pmatrix}2\times0+0\times(-1)\\ 0\times0+1\times(-1)\end{pmatrix}\) | |
| \(=\) | \(\begin{pmatrix}0\\-1\end{pmatrix}\) |
Now map a second point, \((1,1)\):
| \(Q'\) | \(=\) | \(\begin{pmatrix}2&0\\0&1\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}\) |
| \(=\) | \(\begin{pmatrix}2\\1\end{pmatrix}\) |
Find the gradient of the line through \(P'(0,-1)\) and \(Q'(2,1)\):
| \(m\) | \(=\) | \(\dfrac{1-(-1)}{2-0}\) |
| \(=\) | \(\dfrac{2}{2}\) | |
| \(=\) | \(1\) |
Use the image intercept \((0,-1)\) to write the equation:
| \(y\) | \(=\) | \(1\cdot x-1\) |
| \(=\) | \(x-1\) |
The image line is \(y=x-1\).
Write the inverse map. The inverse of the rotation is a \(90^\circ\) clockwise rotation, so \((x,y)=A^{-1}(x',y')\) gives:
| \(x\) | \(=\) | \(y'\) |
| \(y\) | \(=\) | \(-x'\) |
Substitute \(x=y'\) and \(y=-x'\) into \(y=2x+1\):
| \(y\) | \(=\) | \(2x+1\) |
| \(-x'\) | \(=\) | \(2(y')+1\) |
Rearrange to make \(y'\) the subject:
| \(2y'\) | \(=\) | \(-x'-1\) |
| \(y'\) | \(=\) | \(-\dfrac{1}{2}x'-\dfrac{1}{2}\) |
The image line is \(y=-\dfrac{1}{2}x-\dfrac{1}{2}\).
The matrix triples every \(y\)-coordinate, so the inverse map is:
| \(x\) | \(=\) | \(x'\) |
| \(y\) | \(=\) | \(\dfrac{y'}{3}\) |
Substitute into the circle equation \(x^2+y^2=1\):
| \(x^2+y^2\) | \(=\) | \(1\) |
| \((x')^2+\left(\dfrac{y'}{3}\right)^2\) | \(=\) | \(1\) |
| \((x')^2+\dfrac{(y')^2}{9}\) | \(=\) | \(1\) |
The image is the ellipse \(x^2+\dfrac{y^2}{9}=1\).
Apply the matrix to a general point \((x,y)\):
| \(\begin{pmatrix}x'\\y'\end{pmatrix}\) | \(=\) | \(\begin{pmatrix}1&3\\2&6\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\) |
| \(=\) | \(\begin{pmatrix}x+3y\\ 2x+6y\end{pmatrix}\) |
Compare the two image coordinates — the second is twice the first:
| \(y'\) | \(=\) | \(2x+6y\) |
| \(=\) | \(2(x+3y)\) | |
| \(=\) | \(2x'\) |
Every image satisfies \(y=2x\), so the plane collapses onto the line \(y=2x\).
Common pitfalls
Frequently asked questions
How do you find the image of a line under a matrix?
Either map two points on the line and join the images, or substitute the inverse map \((x,y)=A^{-1}(x',y')\) into the equation and simplify.
Why do you use the inverse matrix when finding an image equation?
A point \((x',y')\) is on the image exactly when its pre-image \((x,y)=A^{-1}(x',y')\) lies on the original curve, so you substitute the inverse map into the original equation.
Does a straight line always stay a straight line?
Yes, provided the matrix is non-singular (\(\det A\neq0\)). A singular matrix collapses the whole plane onto a single line through the origin.
What does the determinant tell you about a transformation?
A non-zero determinant means the map is invertible and scales areas by \(|\det A|\). A zero determinant means the map is singular and squashes the plane onto one line.
What is the image of a circle under a dilation?
A dilation stretches a circle into an ellipse. Substituting the inverse map into \(x^2+y^2=r^2\) gives the ellipse equation directly.
Does the order of two transformations matter?
Yes. Doing \(S\) then \(T\) is the single matrix \(TS\), and in general \(TS\neq ST\), so swapping the order can change the image.