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Reflections on a Coordinate Plane

This video is an extension of video mathematics -Transformations: Reflections.

To get a general explanation please watch other videos.

Problem: Reflect the triangle with the vertices of (-1,-2), (-2,1) and (0,3) over the y=x line.

Step 1 Graph vertices and mirror line

Step 2 Find equations of lines perpendicular to mirror the line that passes through each vertex using the point-slope equation.
mirror line y=x
Slope=1
perpendicular lines slope is negative reciprocal. Slope = -1

Point A (-1,-2)	Point B (-2,1)	Point C (0,3)
y+2=-1(x+1)	y-1=-1(x+2)	y-3=-1(x-0)
y+2=-x-1	y-1=-x-2	y-3=-x
y=-x-3	y=-x-1	y=-x+3

Step 3 Solve system of mirror line and perpendicular line to find intersection point on the mirror line.
In all intersections y=x, substitute y=x

Line A y=-x-3	Line B y=-x-1	Line C y=-x+3
x=-x-3	x=-x-1	x=-x+3
2x=-3	2x-1	2x=3
x=-1.5	x=-0.5	x=1.5
y=-1.5	y=-0.5	y=-1.5

Step 4Determine distance between point on mirror line and vertex.
* From now on I’m only going to show calculations for point A. In order to find other mirror vertices, repeat calculations for B and C
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Point A (-1,-2),(-1.5,-1.5)
$d_A=\sqrt{(-1--1.5)^2+(-2--1.5)^2$
$d_A=\sqrt{0.25+0.25}=\sqrt{0.5}$
Step 5 Find point opposite of the initial vertex that lies on the perpendiclarline that has the same distance from the mirror line as to the original vertex.
Point A
$d_A=\sqrt{0.5}=\sqrt{(x--1.5)^2+(y--1.5)^2}$
$0.5=(x--1.5)^2+(y--1.5)^2$
substitute y=-x-3
$0.5=(x+1.5)^2+(-x-3+1.5)^2$
$0.5=x^2+3x+2.25+x^2+3x+2.25$
$0=2x^2+6x+4$
Quadratic formula $\frac{b \pm \sqrt{b^2-4 \cdot a \cdot c}}{2 \cdot a}$
a=2, b=6, c=4
Original Vertex: x=-1 or Mirror Vertex: x=-2
y=-x-3
y=-(-2)-3
y=-1
mirror vertex of A=(-2,-1)