How to reflect and rotate a triangle on a coordinate grid

Key takeaway: This kind of coordinate geometry problem is really about being careful with rules. One transformation changes positions across a mirror line, and the other turns every point around a center by a fixed angle.

(i) Reflection in the line $x=-1$

A reflection in a vertical line keeps the $y$-coordinate the same and changes the $x$-coordinate to the same distance on the other side of the line.

If a point is $(x,y)$, then after reflection in $x=-1$:

$x' = -2 - x, \quad y' = y$

So each vertex of triangle A should be moved the same distance to the other side of $x=-1$.

(ii) Rotation $90^\circ$ clockwise about $(1,-2)$

For a $90^\circ$ clockwise rotation about $(1,-2)$:

1. Translate the point so the center becomes the origin.
2. Rotate using $(u,v) \to (v,-u)$.
3. Translate back.

If a point is $(x,y)$, then relative to $(1,-2)$:

$u=x-1, \quad v=y+2$

After rotation:

$u' = v, \quad v' = -u$

Translate back:

$x' = 1 + v = 1 + (y+2) = y+3$ $y' = -2 + v' = -2 - (x-1) = -x-1$

So the image rule is

$(x,y) \to (y+3,\,-x-1)$

Use that on each vertex of triangle A to plot the rotated image.

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Pitfalls the pros know 👇 For reflections, students often flip across the wrong line because they forget that $x=-1$ is one unit left of the $y$-axis, not the $y$-axis itself.

For rotations, the biggest mistake is rotating around the origin instead of around $(1,-2)$. Always shift to the center first, then rotate, then shift back.

What if the problem changes? If the reflection line were $x=k$ instead of $x=-1$, the reflection rule would become $x' = 2k-x$ while $y$ stays the same. If the rotation were counterclockwise instead of clockwise, the coordinate rule would change to $(x,y) \to (3-y,\,x+? )$ after the same center-shift method, so you would need to redo the mapping carefully.

Tags: reflection, rotation, coordinate geometry