Question

What is the equation of a line that contains the point (1,6) and has a y- intercept of 4

Answer 1

if the y-intercept is 4, namely, the graph intercepts the y-axis at 4, meaning that point is (0, 4).

so this line runs through (1,6) and (0,4).

`\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{6})\qquad \stackrel{y-intercept}{(\stackrel{x_2}{0}~,~\stackrel{y_2}{4})} \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{4-6}{0-1}\implies \cfrac{-2}{-1}\implies 2 \\\\\\ \begin{array}c \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-6=2(x-1) \\\\\\ y-6=2x-2\implies y=2x+4`