To solve the problem of finding the equation of the line passing through (A(1,2)) and forming a triangle (OBC) (with (B) on the x-axis and (C) on the y-axis) of area 1, follow these steps:
Step 1: Define the line equation
Let the line be (y = mx + c).
- (C = (0, c)) (y-intercept), (B = \left(-\frac{c}{m}, 0\right)) (x-intercept).
- Area of (\triangle OBC = \frac{1}{2}|OB \cdot OC| = \frac{1}{2}\left|\left(-\frac{c}{m}\right) \cdot c\right| = 1) → (c^2 = 2|m|).
Step 2: Use point (A(1,2))
Since the line passes through (A):
(2 = m \cdot 1 + c) → (c = 2 - m).
Step 3: Solve for (m) and (c)
Substitute (c = 2 - m) into (c^2 = 2|m|):
- For (m > 0): ((2 - m)^2 = 2m) → (m^2 - 6m + 4 = 0) → (m = 3 \pm \sqrt{5}).
- For (m < 0): No real solutions (discriminant negative).
Step 4: Write the line equations
- If (m = 3 + \sqrt{5}): (c = -1 - \sqrt{5}) → ((3+\sqrt{5})x - y - (1+\sqrt{5}) = 0).
- If (m = 3 - \sqrt{5}): (c = -1 + \sqrt{5}) → ((3-\sqrt{5})x - y + (\sqrt{5}-1) = 0).
Final Answers:
[
\boxed{(3+\sqrt{5})x - y - (1+\sqrt{5}) = 0}
]
[
\boxed{(3-\sqrt{5})x - y + (\sqrt{5}-1) = 0}
]
(Note: Both equations are valid solutions.)
The most concise answer (if only one is expected) is typically the one with the positive y-intercept:
[
\boxed{(3-\sqrt{5})x - y + (\sqrt{5}-1) = 0}
]
But since both are correct, either is acceptable. For simplicity, we present the second one as the main answer.
Answer:
(\boxed{(3 - \sqrt{5})x - y + (\sqrt{5} - 1) = 0})
作者声明:本文包含人工智能生成内容。
