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:
\[ A \ = \ ( x_{ 1 }, \ y_{ 1 } ) \ = \ ( 4, \ 5 ) \]
\[ B \ = \ ( x_{ 2 }, \ y_{ 2 } ) \ = \ ( 9, \ 7 ) \]
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:
\[ c \ = \ 1 \]
\[ m \ = \ 1 \]