Question

Determine the standard form of the equation of the line that passes through ( -5,0) and (0, -9)

2 -9x - 5y = - 45
c. -9x + 5y = 45
2. -92 - 5y = 45
d 5x – 9y = 45
Answers

Answer 1

bearing in mind that the standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

`\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{-9}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-9}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{(-5)}}}\implies \cfrac{-9}{0+5}\implies -\cfrac{9}{5}`

`\bf \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-\stackrel{y_1}{0}=\stackrel{m}{-\cfrac{9}{5}}[x-\stackrel{x_1}{(-5)}]\implies y=-\cfrac{9}{5}(x+5) \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{5}}{5(y) = 5\left[ -\cfrac{9}{5}(x+5) \right]}\implies 5y=-9(x+5) \\\\\\ 5y=-9x-45\implies \blacktriangleright 9x+5y=-45 \blacktriangleleft`