在平面直角坐标系中，函数 ( y = \frac{x+1}{|x|+1} ) 的图像上有三个不同的点...

Mapping out the slope

I'm analyzing the function $y = \frac{x+1}{|x| + 1}$ and focusing on three distinct points that lie on a line with a slope $l$, where the sum of their x-coordinates is zero.

Breaking down the function

OK, let me see: For $x \geq 0$, the function is $y = 1$. For $x < 0$, it simplifies to $y = \frac{x+1}{1 - x}$.

Breaking down the function

I’m analyzing $y = \frac{x+1}{1-x}$ using polynomial division, leading to $y = -1 - \frac{2}{x-1}$ for $x < 0$. This is becoming clearer as I go.

Examining point configurations

I’m thinking through scenarios for three distinct points on a line with slope $l$, where the sum of their x-coordinates is zero. Points can lie in $x > 0$ or $x < 0$.

Revisiting point constraints

OK, let me think: for points with $x > 0$, the sum of their x-coordinates can't be zero. This forces me to consider points with non-positive x-coordinates.

Mapping out distinct points

I'm curious about two points in $x \geq 0$ and one in $x < 0$. The third point is $(-(a+b), \frac{1 - (a+b)}{1 + a + b})$, ensuring distinctness.

Noticing a contradiction

OK, let's think it through: Because $(a,1)$ and $(b,1)$ have a slope of 0, having the third point also with a y-value of 1 just isn't feasible.

Evaluating collinearity

I’m examining the scenario with two points in $x \geq 0$ and one in $x < 0$. It’s interesting to see that maintaining slope consistency requires the third point to have a y-value of 1.

Analyzing slopes

I'm thinking through one point in $x \geq 0$ and two in $x < 0$, ensuring distinctness and maintaining distinct slope angles. This involves confirming the slope distance from each to the others.

Piecing together collinear slopes

I'm calculating slopes from $(a,1)$ to $(x1,y1)$ and $(x2,y2)$, aiming for identical slopes to achieve collinearity. This involves algebraic simplification and slope equality.

Mapping out solutions

I’m working through parametric symmetry or setting $a=0$. I find that $a=0$ doesn't fit, so I’m focusing on $a>0$ to pinpoint potential points.

Taking a closer look

When both $x_1$ and $x_2$ are negative, the sum is negative, which can't be zero. Thus, we must have one or two points in $x > 0$.

Establishing the equation

I’m working through slopes to establish relationships among $a$, $x_1$, and $l$, then using $x_1 + x_2 = -a$ to find $x_2$.

Ensuring slopes align

I’m confirming slopes for $(x_1, y_1)$ and $(x_2, y_2)$ match. With $x_1$ between $-a$ and 0, $x_2 = -a - x_1$ also falls in this range.

Aligning slopes

I’m comparing slopes from $(x_1, y_1)$ to $(x_2, y_2)$ with $(a,1)$ using Python. Adjusting $a>0$ and $x_1$ within $(-a,0)$, I’ll identify overlapping slopes for consistency.

Examining slope trends

I'm thinking through how slopes change with $a$ from 0 to infinity, using a Python numeric approach to find a range of slopes consistent with collinearity.

Ensuring point uniqueness

I'm checking that $x_1 \neq a$ and $x_2 \neq a$, since both are negative and $a > 0$. Maintaining $x_1 \neq x_2$ only causes an issue when $a = 0$.

Confirming collinearity

I’m focusing on a numeric approach to verify collinearity by comparing slopes with Python, ensuring all points align consistently within the given range.

推理关于斜率取值范围 for 48 seconds