How do you write an equation in standard form for a line with a slope of 1/4 and goes through point (4, 0)?

Dymnembalmese2n
2022-09-21
Answered

You can still ask an expert for help

Kelbelol

Answered 2022-09-22
Author has **10** answers

the equation of a line in standard form is.

$\overline{\underline{\left|{\frac{2}{2}}{Ax+By=C}{\frac{2}{2}}\right|}}$

where A is a positive integer and B, C are integers

to begin express the equation in point-slope form

$\u2022{x}y-{y}_{1}=m(x-{x}_{1})$

$\text{where m is the slope and}\phantom{\rule{1ex}{0ex}}({x}_{1},{y}_{1})\phantom{\rule{1ex}{0ex}}\text{a point on the line}$

$\text{here}\phantom{\rule{1ex}{0ex}}m=\frac{1}{4}\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}({x}_{1},{y}_{1})=(4,0)$

$\Rightarrow y=\frac{1}{4}(x-4)$

$\Rightarrow y=\frac{1}{4}x-1$

multiply all terms by 4

$\Rightarrow 4y=x-4$

$\Rightarrow x-4y=4\leftarrow {\text{in standard form}}$

$\overline{\underline{\left|{\frac{2}{2}}{Ax+By=C}{\frac{2}{2}}\right|}}$

where A is a positive integer and B, C are integers

to begin express the equation in point-slope form

$\u2022{x}y-{y}_{1}=m(x-{x}_{1})$

$\text{where m is the slope and}\phantom{\rule{1ex}{0ex}}({x}_{1},{y}_{1})\phantom{\rule{1ex}{0ex}}\text{a point on the line}$

$\text{here}\phantom{\rule{1ex}{0ex}}m=\frac{1}{4}\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}({x}_{1},{y}_{1})=(4,0)$

$\Rightarrow y=\frac{1}{4}(x-4)$

$\Rightarrow y=\frac{1}{4}x-1$

multiply all terms by 4

$\Rightarrow 4y=x-4$

$\Rightarrow x-4y=4\leftarrow {\text{in standard form}}$

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