• Answer:
[tex] \Large{\boxed{\sf y = \dfrac{1}{2}x}} [/tex]
[tex] \\ [/tex]
• Explanation:
We know that the equation of a line in slope-intercept form is:
[tex] \Large{\left[ \begin{array}{c c c} \underline{\tt Slope-Intercept \ Form \text{:}} \\ ~ \\ \tt y = mx + b \end{array} \right] } [/tex]
Where:
• (x , y) is a point on the line.
• m is the slope of the line.
• b is the y-intercept.
[tex] \\ [/tex]
Since we are given the value of the slope, we can substitute it into the equation:
[tex] \sf y = \dfrac{1}{2}x + b [/tex]
[tex] \\ [/tex]
We know that the line passes through (4 , 2), which means that the coordinates of this point verify the equation of the line. Therefore, we can substitute its coordinates into the equation and solve for b:
[tex] \sf (\overbrace{\sf 4}^{\sf x} \ , \ \underbrace{\sf 2}_{\sf y}) \\ \\ \sf \longrightarrow 2 = \dfrac{1}{2}(4) + b \\ \\ \longrightarrow \sf 2 = 2 + b \\ \\ \longrightarrow \sf 2 - 2 = 2 + b - 2 \\ \\ \boxed{\boxed{\sf b = 0}}[/tex]
[tex] \\ [/tex]
Therefore, the equation of the line is:
[tex] \boxed{\boxed{\sf y = \dfrac{1}{2}x }} [/tex]
