Question: Graph at top-left: a coordinate-plane sketch with two intersecting lines, one labeled $y = -x+1$ (descending) and one labeled $y = x-1$ (ascending), with point $A(2,1)$ marked and a reflected point near $A'(0,1)$.
$A(2,1)$, $y = -x+1$ -> $y=mx+b$
$m_1 = -1$ $m = \text{rise}/\text{run} = -1/1$ $b = 1$
line perpendicular to the line of reflection
$m_1,m_2 = -1$
$m_2 = 1$
y = m_2x + b$
$1 = 1(2)+b$
$1 = 2+b$
$1-2=b$
$-1=b$
y = x-1$
find the midpoint $(x_m, y_m)$
$y_1 = y_2$
$-x+1 = x-1$
$1+1 = x+x$
$\text{divide } 2/2 = 2x/2$
$1 = x$
$11 = x_m$
y=x-1$
$y_m = 0$
$(x_m,y_m) = (1,0)$
$x_m = \frac{x_1+x_2}{2}$ $y_m = \frac{y_1+y_2}{2}$
$1 = \frac{2+x_2}{2}$
$2 = 2+x_2$
$2-2 = x_2$
$0 = x_2$
$0 = 1+y_2$
$0 = 1+\frac{y_2}{2}$
$-1 = y_2$
$A(2,1) \to A'(0,1)$
General formula
$(x,y) \to \left[\frac{(b^2 - a^2)x - 2ab y - 2ac}{a^2 + b^2}, \frac{(a^2 - b^2)y - 2ab x - 2bc}{a^2 + b^2}\right]$
$A(2,1)$ about $y=-x+1$ -> $ax+by+c = 0$, $a=1$, $b=1$, $c=-1$ $-x+y-1=0$
$A(2,1) \to \left[\frac{(1^2 - 1^2)(2) - 2(1)(1)(1) - 2(1)(-1)}{1^2 + 1^2}, \frac{(1^2 - 1^2)(1) - 2(1)(1)(2) - 2(1)(-1)}{1^2 + 1^2}\right]$
$\to \left(\frac{0 - 2 + 2}{2}, \frac{0 - 4 + 2}{2}\right)$
$\to \left(\frac{0}{2}, \frac{-2}{2}\right)$
$\to (0, -1)$ final answer
Solve again using these methods
1. **State the problem:** Reflect point $A(2,1)$ about the line $y = -x + 1$ and find the coordinates of the reflected point $A'$.
2. **Identify the line equation in standard form:** The line is $y = -x + 1$, rewrite as $x + y - 1 = 0$ where $a=1$, $b=1$, $c=-1$.
3. **Recall the reflection formula:** For a point $(x,y)$ reflected about line $ax + by + c = 0$, the reflected point $(x', y')$ is given by
$$x' = \frac{(b^2 - a^2)x - 2ab y - 2ac}{a^2 + b^2}, \quad y' = \frac{(a^2 - b^2)y - 2ab x - 2bc}{a^2 + b^2}$$
4. **Plug in values:**
$a=1$, $b=1$, $c=-1$, $x=2$, $y=1$. Calculate numerator and denominator for $x'$:
$$ (1^2 - 1^2) \cdot 2 - 2 \cdot 1 \cdot 1 \cdot 1 - 2 \cdot 1 \cdot (-1) = 0 - 2 + 2 = 0 $$
Denominator:
$$1^2 + 1^2 = 1 + 1 = 2$$
So,
$$x' = \frac{0}{2} = 0$$
5. **Calculate numerator for $y'$:**
$$ (1^2 - 1^2) \cdot 1 - 2 \cdot 1 \cdot 1 \cdot 2 - 2 \cdot 1 \cdot (-1) = 0 - 4 + 2 = -2 $$
Denominator same as before:
$$2$$
So,
$$y' = \frac{-2}{2} = -1$$
6. **Final reflected point:**
$$A' = (0, -1)$$
7. **Summary:** The reflection of $A(2,1)$ about the line $y = -x + 1$ is $A'(0,-1)$, confirming the calculation using the reflection formula step-by-step.