Question

1. describe your process for finding slope. why does it not matter which points you choose to find it?

now, consider the algebraic process for finding slope and use it to inform your responses.

1. algebraically show why your response to question 2 is true.

Explanation:

Question 2

Step1: Recall Slope Formula

The slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \( m=\frac{y_2 - y_1}{x_2 - x_1} \), where \( x_2
eq x_1 \). To find the slope, identify two points on the line, subtract their \( y \)-coordinates (rise) and \( x \)-coordinates (run), then divide rise by run.

Step2: Explain Invariance

A line has a constant slope (it's straight, so steepness doesn't change). Any two points on the line will give the same ratio of vertical change to horizontal change because the line's direction is uniform.

Question 3

Step1: Let Points on Line

Let a line have three points: \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \), all lying on the same line.

Step2: Slope Between \( A \) and \( B \)

Slope \( m_{AB}=\frac{y_2 - y_1}{x_2 - x_1} \).

Step3: Slope Between \( B \) and \( C \)

Slope \( m_{BC}=\frac{y_3 - y_2}{x_3 - x_2} \).

Step4: Use Linearity (Equation of Line)

The equation of a line is \( y = mx + b \). For point \( A \): \( y_1=mx_1 + b \); for \( B \): \( y_2=mx_2 + b \); for \( C \): \( y_3=mx_3 + b \).
Subtract \( y_1 \) from \( y_2 \): \( y_2 - y_1=m(x_2 - x_1) \), so \( m=\frac{y_2 - y_1}{x_2 - x_1} \).
Subtract \( y_2 \) from \( y_3 \): \( y_3 - y_2=m(x_3 - x_2) \), so \( m=\frac{y_3 - y_2}{x_3 - x_2} \).
Thus, \( m_{AB}=m_{BC}=m \), proving slope is the same for any two points.

Question 2 Answer:

To find slope, use \( m = \frac{y_2 - y_1}{x_2 - x_1} \) for two points \((x_1,y_1)\), \((x_2,y_2)\) on the line (calculate rise over run). It doesn't matter which points you choose because a line has constant steepness (uniform slope), so any two points give the same rise - over - run ratio.

Question 3 Answer:

Let a line have points \( (x_1,y_1) \), \( (x_2,y_2) \), \( (x_3,y_3) \) on it. The line's equation is \( y=mx + b \), so \( y_1=mx_1 + b \), \( y_2=mx_2 + b \), \( y_3=mx_3 + b \). Subtracting, \( y_2 - y_1=m(x_2 - x_1)\Rightarrow m=\frac{y_2 - y_1}{x_2 - x_1} \), and \( y_3 - y_2=m(x_3 - x_2)\Rightarrow m=\frac{y_3 - y_2}{x_3 - x_2} \). Thus, slope between any two points is equal (constant for the line).

Answer:

Step1: Let Points on Line

Let a line have three points: \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \), all lying on the same line.

Step2: Slope Between \( A \) and \( B \)

Slope \( m_{AB}=\frac{y_2 - y_1}{x_2 - x_1} \).

Step3: Slope Between \( B \) and \( C \)

Slope \( m_{BC}=\frac{y_3 - y_2}{x_3 - x_2} \).

Step4: Use Linearity (Equation of Line)

The equation of a line is \( y = mx + b \). For point \( A \): \( y_1=mx_1 + b \); for \( B \): \( y_2=mx_2 + b \); for \( C \): \( y_3=mx_3 + b \).
Subtract \( y_1 \) from \( y_2 \): \( y_2 - y_1=m(x_2 - x_1) \), so \( m=\frac{y_2 - y_1}{x_2 - x_1} \).
Subtract \( y_2 \) from \( y_3 \): \( y_3 - y_2=m(x_3 - x_2) \), so \( m=\frac{y_3 - y_2}{x_3 - x_2} \).
Thus, \( m_{AB}=m_{BC}=m \), proving slope is the same for any two points.

Question 2 Answer:

To find slope, use \( m = \frac{y_2 - y_1}{x_2 - x_1} \) for two points \((x_1,y_1)\), \((x_2,y_2)\) on the line (calculate rise over run). It doesn't matter which points you choose because a line has constant steepness (uniform slope), so any two points give the same rise - over - run ratio.

Question 3 Answer:

Let a line have points \( (x_1,y_1) \), \( (x_2,y_2) \), \( (x_3,y_3) \) on it. The line's equation is \( y=mx + b \), so \( y_1=mx_1 + b \), \( y_2=mx_2 + b \), \( y_3=mx_3 + b \). Subtracting, \( y_2 - y_1=m(x_2 - x_1)\Rightarrow m=\frac{y_2 - y_1}{x_2 - x_1} \), and \( y_3 - y_2=m(x_3 - x_2)\Rightarrow m=\frac{y_3 - y_2}{x_3 - x_2} \). Thus, slope between any two points is equal (constant for the line).