Question: General formula
(x, y) -> [\frac{(b^2 - a^2)x - 2aby - 2acx - 2bc}{a^2 + b^2}]
A(2,1) -> [\frac{(1 - 1^2)(2) - 2(1)(1)(1) - 2(1)(1)(2) - 2(1)(1)(1)}{1^2 + 1^2}]
-> [\frac{(0 - 2 + 2)}{2} , \frac{(0 - 4 + 2)}{2}]
-> (0, -1)
A(2,1) about y = -x + 1 -> ax + by + c = 0, a=1, b=1, c=-1, -x + y - 1 = 0
A(2,1) -> [\frac{(1 - 1^2)(1) - 2(1)(1)(1) - 2(1)(1)(2) - 2(1)(1)(1)}{1^2 + 1^2}]
-> [\frac{(0 - 2 + 2)}{2} , \frac{(0 - 4 + 2)}{2}]
-> (0, -1)
d = \frac{ax + by + c}{a^2 + b^2} = \frac{1(2) + 1(1) + (-1)}{1^2 + 1^2} = \frac{2 + 1 - 1}{2} = \frac{2}{2} = 1
final answer
(x', y') -> (x - 2ad, y - 2bd) = (2 - 2(1)(1), 1 - 2(1)(1)) = (0, -1)
1. **State the problem:** We want to find the reflection of point $A(2,1)$ about the line $y = -x + 1$.
2. **Rewrite the line in standard form:** The line $y = -x + 1$ can be written as $ax + by + c = 0$ where $a = 1$, $b = 1$, and $c = -1$.
3. **Calculate the distance $d$ from point $A$ to the line:**
$$
d = \frac{|ax + by + c|}{\sqrt{a^2 + b^2}} = \frac{|1(2) + 1(1) - 1|}{\sqrt{1^2 + 1^2}} = \frac{|2 + 1 - 1|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}
$$
4. **Use the reflection formula:** The reflection $(x', y')$ of point $(x, y)$ about the line $ax + by + c = 0$ is given by
$$
(x', y') = \left(x - \frac{2a(ax + by + c)}{a^2 + b^2}, y - \frac{2b(ax + by + c)}{a^2 + b^2}\right)
$$
5. **Calculate the numerator $ax + by + c$ for point $A$:**
$$
ax + by + c = 1(2) + 1(1) - 1 = 2 + 1 - 1 = 2
$$
6. **Calculate the denominator $a^2 + b^2$:**
$$
1^2 + 1^2 = 1 + 1 = 2
$$
7. **Calculate the reflected coordinates:**
$$
x' = 2 - \frac{2 \times 1 \times 2}{2} = 2 - 2 = 0
$$
$$
y' = 1 - \frac{2 \times 1 \times 2}{2} = 1 - 2 = -1
$$
8. **Final answer:** The reflection of point $A(2,1)$ about the line $y = -x + 1$ is
$$
(0, -1)
$$