Question

Reflect the figure over the line y = x + 2.

Answer 1

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:

`u=\frac{(1-m^2)^{}x+2my-2mb}{m^2+1}`

and

`v=\frac{(m^2-1)^{}y+2mx+2b}{m^2+1}`

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):

`\begin{gathered} u=\frac{(1-m^2)^{}x+2my-2mb}{m^2+1} \\ u=(0+2(1)(4)-2(1)(2))/(2) \\ u=2 \\ \text{and} \\ v=\frac{(m^2-1)^{}y+2mx+2b}{m^2+1} \\ v=(0+2(1)(-4)+2(2))/(2) \\ v=-2 \end{gathered}`

reflecting (-9,6):

`\begin{gathered} u=\frac{(1-m^2)^{}x+2my-2mb}{m^2+1} \\ u=(0+2(1)(6)-2(1)(2))/(2) \\ u=4 \\ \text{and} \\ v=((m^2-1)y+2mx+2b)/(m^2+1) \\ v=(0+2(1)(-9)+2(2))/(2) \\ v=-7 \end{gathered}`

reflecting (-8,9):

`\begin{gathered} u=\frac{(1-m^2)^{}x+2my-2mb}{m^2+1} \\ u=(0+2(1)(9)-2(1)(2))/(2) \\ u=7 \\ \text{and} \\ v=((m^2-1)y+2mx+2b)/(m^2+1) \\ v=(0+2(1)(-8)+2(2))/(2) \\ v=-6 \end{gathered}`

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: