Question

A line contains the points (4,2) and (0,1). What is the equation of the line?

Answer 1

$\bf (\stackrel{x_1}{4}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{2}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{4}}}\implies \cfrac{-1}{-4}\implies \cfrac{1}{4}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{2}=\stackrel{m}{\cfrac{1}{4}}(x-\stackrel{x_1}{4}) \\\\\\ y-2=\cfrac{1}{4}x-1\implies y = \cfrac{1}{4}x+1$

Answer 2

Answer:

y = 1/4x + 1

Step-by-step explanation:

(0,1) = y's intersections is 1.

therefore, y = ?x +1

If you put (4,2) you can figure out the coefficient  of x.

2 = (4 x ?) +1

?(x's coefficient)  = 1/4

therefore, the equation is,

y = 1/4x + 1