The triangle shown has vertices:
(-4, 4)
(-9, 6)
(-8, 9)
We need to reflect these 3 points [vertices of the triangle] about the line y = x + 2.
If the reflected point(s) are of the form (u, v), then we can use the formula shown below to find the points:
[tex]u=\frac{(1-m^2)^{}x+2my-2mb}{m^2+1}[/tex]
and
[tex]v=\frac{(m^2-1)^{}y+2mx+2b}{m^2+1}[/tex]
Where
x and y are the coordinates of the points we are reflecting and m is the slope fo the line and b is the y-intercept of the line about which we are making the reflection
y = x + 2
y = mx + b [slope intercept form of line]
Hence,
m = 1
b = 2
Now,
reflecting (-4,4):
[tex]\begin{gathered} u=\frac{(1-m^2)^{}x+2my-2mb}{m^2+1} \\ u=\frac{0+2(1)(4)-2(1)(2)}{2} \\ u=2 \\ \text{and} \\ v=\frac{(m^2-1)^{}y+2mx+2b}{m^2+1} \\ v=\frac{0+2(1)(-4)+2(2)}{2} \\ v=-2 \end{gathered}[/tex]
reflecting (-9,6):
[tex]\begin{gathered} u=\frac{(1-m^2)^{}x+2my-2mb}{m^2+1} \\ u=\frac{0+2(1)(6)-2(1)(2)}{2} \\ u=4 \\ \text{and} \\ v=\frac{(m^2-1)y+2mx+2b}{m^2+1} \\ v=\frac{0+2(1)(-9)+2(2)}{2} \\ v=-7 \end{gathered}[/tex]
reflecting (-8,9):
[tex]\begin{gathered} u=\frac{(1-m^2)^{}x+2my-2mb}{m^2+1} \\ u=\frac{0+2(1)(9)-2(1)(2)}{2} \\ u=7 \\ \text{and} \\ v=\frac{(m^2-1)y+2mx+2b}{m^2+1} \\ v=\frac{0+2(1)(-8)+2(2)}{2} \\ v=-6 \end{gathered}[/tex]
Thus, the points and their reflections are:
(-4, 4) >>> (2, -2)(-9, 6) >>> (4, -7)(-8, 9) >>> (7, -6)
Graph these points are connect as a triangle.
Shown below:
