The** aim of this question** is to understand and apply **different forms of line**. It also covers **different parameters** used in the linear equations such as **slope and intercepts.**

According to** two point form**, an equation can be written in the following form:

\[ \dfrac{ y – y_{ 1 } }{ y_{ 2 } – y_{ 1 } } \ = \ \dfrac{ x – x_{ 1 } }{ x_{ 2 } – x_{ 1 } } \]

Where $ ( x_{ 1 }, \ y_{ 1 } ) $ and $ ( x_{ 2 }, \ y_{ 2 } ) $ are any **two points lying on the line**. According to **slope intercept form**, an equation can be written in the following form:

\[ y \ = \ m x + c \]

Where $ m $ and $ c $ are the **slope and y-intercept,** respectively.

## Expert Answer

**Given **that there are **two points:**

\[ A \ = \ ( x_{ 1 }, \ y_{ 1 } ) \ = \ ( 4, \ 5 ) \]

\[ B \ = \ ( x_{ 2 }, \ y_{ 2 } )Â \ = \ ( 9, \ 7 ) \]

**This implies that:**

\[ x_{ 1 } \ = \ 4 \]

\[ x_{ 2 } \ = \ 9 \]

\[ y_{ 1 } \ = \ 5 \]

\[ y_{ 2 } \ = \ 7 \]

According to the **two-point form** of a line:

\[ \dfrac{ y – y_{ 1 } }{ y_{ 2 } – y_{ 1 } } \ = \ \dfrac{ x – x_{ 1 } }{ x_{ 2 } – x_{ 1 } } \]

**Substituting values:**

\[ \dfrac{ y – 5 }{ 7 – 5 } \ = \ \dfrac{ x – 4 }{ 9 – 4 } \]

\[ \dfrac{ y – 5 }{ 2 } \ = \ \dfrac{ x – 4 }{ 5 } \]

\[ 5 ( y – 5 ) \ = \ 2 ( x – 4 ) \]

\[ 5 y – 25 \ = \ 2 x – 8 \]

\[ 5 y \ = \ 2 x – 8 + 25 \]

\[ 5 y \ = \ 2 x + 17 \]

\[ y \ = \ \dfrac{ 2 }{ 5 } x + \dfrac{ 17 }{ 5 } \]

Comparing the above equation with the following** slope intercept form** of a line:

\[ y \ = \ m x + c \]

We can **conclude** that:

\[ c \ = \ \dfrac{ 17 }{ 5 } \]

\[ m \ = \ \dfrac{ 2 }{ 5 } \]

Which is the **slope of the given line.**

## Numerical Result

\[ m \ = \ \dfrac{ 2 }{ 5 } \]

## Example

**Given following points, find the slope and intercept of the line joining these two points:**

\[ A \ = \ ( 1, \ 2 ) \]

\[ B \ = \ ( 3, \ 4 ) \]

**Here:**

\[ x_{ 1 } \ = \ 1 \]

\[ x_{ 2 } \ = \ 3 \]

\[ y_{ 1 } \ = \ 2 \]

\[ y_{ 2 } \ = \ 4 \]

According to the **two-point form** of a line:

\[ \dfrac{ y – y_{ 1 } }{ y_{ 2 } – y_{ 1 } } \ = \ \dfrac{ x – x_{ 1 } }{ x_{ 2 } – x_{ 1 } } \]

**Substituting values:**

\[ \dfrac{ y – 2 }{ 4 – 2 } \ = \ \dfrac{ x – 1 }{ 3 – 1 } \]

\[ \dfrac{ y – 2 }{ 2 } \ = \ \dfrac{ x – 1 }{ 2 } \]

\[ y – 2 \ = \ x – 1 \]

\[ y \ = \ x – 1 + 2 \]

\[ y \ = \ x + 1 \]

Comparing the above equation with the following **slope intercept** form of a line:

\[ y \ = \ m x + c \]

We can **conclude** that:

\[ c \ = \ 1 \]

\[ m \ = \ 1 \]