Reflection across a line: coordinates and image equations
Reflect points and entire figures across the coordinate axes or an arbitrary line, using projections and inverse substitution.
Content updated: October 22, 2026
Reflection across a line fixes every point on that line. For a point off the line, its projection onto the line is the midpoint of the point and its image. This gives both a construction and a coordinate formula.
Learning goals: find images across the coordinate axes and an arbitrary line, and write image equations of lines, circles, and parabolas.
Prerequisites: projection and distance from a point to a line (English lesson); inverse substitution (English lesson).
On this page: The idea · Method · Examples · Practice · Further learning
1. Definition and formula
Let the axis be $d:ax+by+c=0$, with $a^2+b^2>0$. The reflection $D_d$ fixes every point on $d$. For $M$ outside $d$, it maps $M$ to $M^{\prime}$ so that $d$ is the perpendicular bisector of $MM^{\prime}$. If $H$ is the projection of $M$ onto $d$, then $M^{\prime}=2H-M$. This formula also holds when $M$ lies on $d$, in which case $H=M=M^{\prime}$; these coincident points must not be used to define a line.
Set $q=ax+by+c$ and $s=a^2+b^2$. Since $H=M-\frac q s(a;b)$, the image coordinates are
In particular, $D_{Ox}(x;y)=(x;-y)$ and $D_{Oy}(x;y)=(-x;y)$. Applying $D_d$ twice returns the original point. Decomposing a vector into components along and perpendicular to the axis, reflection preserves the first component and negates the second; its squared length is therefore unchanged. Thus reflection preserves distances and maps triangles to congruent triangles, lines to lines, and circles to circles of the same radius.
2. Finding an image equation
For the image of a line, reflect two distinct points. If the line meets the axis, their intersection is a convenient fixed point. If it is parallel to the axis, reflect one point and retain the direction parallel to the axis; do not assume that an intersection exists in this case.
For a figure $F(x,y)=0$, express the original point $M$ in terms of the image point $(X;Y)$ using the same reflection formula, since $D_d^{-1}=D_d$. Substitution into $F=0$ gives the image equation. The converse holds because reflecting again returns the original point; a single inclusion is not enough.
3. Six worked examples
Example 1. Images across Ox and Oy
For $A(1;-2)$, reflection across $Ox$ gives $A_1(1;2)$, and across $Oy$ gives $A_2(-1;-2)$. For $B(3;-1)$, the corresponding images are $B_1(3;1)$ and $B_2(-3;-1)$.
Example 2. Reflecting a line across the coordinate axes
Let $\Delta:x+3y-1=0$. Across $Ox$, the preimage of $(X;Y)$ is $(X;-Y)$, so $\Delta_1:x-3y-1=0$. Across $Oy$, the preimage is $(-X;Y)$, so $\Delta_2:-x+3y-1=0$.
Example 3. A point and an arbitrary reflection axis
Let $M(6;1)$ and $d:2x-y-1=0$. Write its projection as $H(m;2m-1)$. The vector $\overrightarrow{MH}=(m-6;2m-2)$ is perpendicular to direction vector $(1;2)$, so $m-6+2(2m-2)=0$. Hence $m=2$ and $H(2;3)$. Reflecting about this midpoint gives
Check: the midpoint of $(6;1)$ and $(-2;5)$ is $(2;3)$, which lies on $d$; their joining vector is $(-8;4)=-4(2;-1)$, perpendicular to $d$.
Example 4. Two methods for the image of a line
Let $\Delta:x+y-2=0$ and axis $d:x-2y+1=0$. Two-point method: the intersection $M(1;1)$ is fixed. For $A(2;0)$, we have $q=3$ and $s=5$, so $A^{\prime}(4/5;12/5)$. The two images are distinct because $A\ne M$. Their joining line is $\Delta^{\prime}:7x+y-8=0$.
Inverse substitution: the preimage of $(X;Y)$ across the same axis has coordinates
Substitution into $x+y-2=0$ gives $(7X+Y-8)/5=0$, again yielding $7x+y-8=0$. This formula also covers fixed points without attempting to construct a line through coincident points.
Example 5. The image of a circle
Let $(C):(x-3)^2+(y+1)^2=4$ and axis $x-1=0$. Its center $I(3;-1)$ maps to $I^{\prime}(-1;-1)$ because their midpoint has x-coordinate $1$. The radius remains $2$, so $(C^{\prime}):(x+1)^2+(y+1)^2=4$.
Example 6. The image of a parabola
Let $(P):y=x^2+1$ and axis $x-2y+1=0$. Using the inverse-substitution formulas from Example 4,
Expanding, rearranging, and renaming the coordinates gives
This is still a parabola, since reflection preserves shape and has an inverse. Its axis has changed direction, so its equation may contain an $xy$ term and need not be a graph of the form $y=f(x)$.
4. Practice exercises
- Find the image of $P(3;1)$ across the axis $y=x$.
- Find the image of $x=3$ across the axis $x=1$.
- Find the image of $y=x^2$ across the axis $Ox$.
Show answers and hints
- The axis $x-y=0$ gives $q=2$ and $s=2$, so $P^{\prime}(1;3)$.
- The new x-coordinate is $2-3=-1$, so the image is $x=-1$.
- Substituting the preimage $(X;-Y)$ gives $-Y=X^2$, so $y=-x^2$.
5. Common mistakes
- Changing only one coordinate when the axis is neither $Ox$ nor $Oy$.
- Forgetting the factor $2$ when passing from the projection to the reflected point.
- Using the joining line of a point and its image when the point lies on the reflection axis.
- Rejecting a parabola equation merely because it contains an $xy$ term.
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