Question

What is the equation in standard form of the line shown on the graph?

Answer 1

Answer:  The required equation of the graphed line is $3x-4y=12.$

Step-by-step explanation:  We are given to find the equation in standard form of the line shown on the graph.

From the graph, we note that

the line passes through the points (4, 0) and (0, -3).

We know that

the slope of a line passing through the points (a, b) and (c, d) is given by

$m=\dfrac{d-b}{c-a}.$

So, the slope of he graphed line is given by

$m=\dfrac{-3-0}{0-4}=\dfrac{-3}{-4}=\dfrac{3}{4}.$

Since (4, 0) is one of the points on the line, so the equation of the graphed line will be

$y-0=m(x-4)\\\\\Rightarrow y=\dfrac{3}{4}(x-4)\\\\\\\Rightarrow 4y=3x-12\\\\\Rightarrow 3x-4y=12.$

Thus, the required equation of the graphed line is $3x-4y=12.$

Answer 2

well, let's check the graph, and use two points off of it..hmmmm it passes through (4, 0) and it also passes through (0, -4).

so, what would be the equation of a line that passes through those two points?

keeping in mind that

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

$\bf (\stackrel{x_1}{4}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{-4}) ~\hfill slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-4-0}{0-4}\implies 1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-0=1(x-4)\implies y=x-4 \\\\\\ -x+y=-4\implies \stackrel{\textit{standard form}}{x-y=4}\\\hspace{34em}$