直线 $2x + y = 1$ 与曲线 $x^2 - xy - y^2 + 5 = 0$ 相交于点 $A$ 和 $B$。 证明长度 $AB$ 为 $5\sqrt{5}$ 单位。

1. Express $y$ in terms of $x$ from the linear equation

From $2x + y = 1$, we have:
$y = 1 - 2x$

2. Substitute the expression for $y$ into the quadratic equation

Substitute $y = 1 - 2x$ into $x^2 - xy - y^2 + 5 = 0$:
$x^2 - x(1 - 2x) - (1 - 2x)^2 + 5 = 0$

3. Expand and simplify the resulting quadratic equation in $x$

$x^2 - x + 2x^2 - (1 - 4x + 4x^2) + 5 = 0$
$3x^2 - x - 1 + 4x - 4x^2 + 5 = 0$
$-x^2 + 3x + 4 = 0$
$x^2 - 3x - 4 = 0$

4. Solve the quadratic equation for $x$

$(x - 4)(x + 1) = 0$
$x = 4$ or $x = -1$

5. Find the corresponding $y$-coordinates

When $x = 4$, $y = 1 - 2(4) = -7$
When $x = -1$, $y = 1 - 2(-1) = 3$
The points of intersection are $A(4, -7)$ and $B(-1, 3)$.

6. Calculate the length of the line segment $AB$

Length $AB = \sqrt{(4 - (-1))^2 + (-7 - 3)^2}$
$= \sqrt{5^2 + (-10)^2}$
$= \sqrt{25 + 100}$
$= \sqrt{125}$
$= \sqrt{25 \times 5}$
$= 5\sqrt{5}$ units