Question

What is the equation in standard form of the line that passes through the point (1,24) and has a slope of negative 0.6

Answer 1

let's first off convert the 0.6 to a fraction, and then let's keep in mind that

Standard Form of a Linear Equation

- variables must be on the left-hand-side, usually sorted in ascending order

- there must not be any fractions, just integers

- the variable "x" must not have a negative coefficient.

`\bf 0.\underline{6}\implies \cfrac{06}{1\underline{0}}\implies \cfrac{3}{5}\impliedby m = slope ~\hspace{12em} (\stackrel{x_1}{1}~,~\stackrel{y_1}{24}) \\\\[-0.35em] ~\dotfill\\\\ \begin{array}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-24=\cfrac{3}{5}(x-1)\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{5}}{5y-120=3(x-1)} \\\\\\ 5y-120=3x-3\implies -3x+5y=117\implies \stackrel{\textit{standard form}}{3x-5y=-117}`