What is the general form of the equation of the line shown?

Mathematics

2 answers:

Mrac [35]3 years ago

7 0

Answer:

B. x - y = 0

Step-by-step explanation:

ZanzabumX [31]3 years ago

5 0

Check the picture below. So let's use those two points on the line.

$\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{3}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{3-0}{3-0}\implies \cfrac{3}{3}\implies 1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-0=1(x-0)\implies y=x \\\\\\ -x+y=0\implies \stackrel{\textit{standard form}}{x-y=0}$

bearing in mind that the standard form is also a general form.

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

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