The linear equality represented by the graph is \(\boxed{y\leqslant\frac{1}{3}x-1.3}\).
Further explanation:
It is given that a line passes through points \(\left({0,-\,1.3}\right)\) and \(\left({3,-\,0.3}\right)\) as shown below in Figure 1.
The slope of a line passes through points \(\left({{x_1},{y_1}}\right)\) and \(\left({{x_2},{y_2}}\right)\) is calculated as follows:
\(m=\frac{{{y_2}-{y_1}}}{{{x_2}-{x_1}}}\) .......(1)
Here, the slope of a line is denoted as \(m\) and points are \(\left({{x_1},{y_1}}\right)\) and \(\left({{x_2},{y_2}}\right)\) .
Substitute \(0\) for \({x_1}\) , \(-1.3\) for \({y_1}\) , \(3\) for \({x_2}\) and \(-0.3\) for \({y_2}\) in equation (1) to obtain the slope of a line that passes through points \(\left({0,-\,1.3}\right)\) and \(\left({3,-\,0.3}\right)\) .
\(\begin{aligned}m&=\frac{{-0.3-\left({-1.3}\right)}}{{3-0}}\\&=\frac{1}{3}\\\end{aligned}\)
Therefore, the slope is \(\frac{1}{3}\) .
The point-slope form of the equation of a line with slope \(m\) passes through point \(\left({{x_1},{y_1}}\right)\) is represented as follows:
\(y-{y_1}=m\left({x-{x_1}}\right)\) ......(2)
Substitute \(0\) for \({x_1}\) , \(-1.3\) for \({y_1}\) and \(\frac{1}{3}\) for \(m\) in equation (2) to obtain the equation of line.
\(\begin{aligned}y-\left({-1.3}\right)&=\frac{1}{3}\left({x-0}\right)\\y+1.3&=\frac{1}{3}x\\y&=\frac{1}{3}x-1.3\\\end{aligned}\)
Therefore, the value of \(y\) is \(\frac{1}{3}x-1.3\) .
Since the shaded part in Figure 1 is below the equation of line \(y=\frac{1}{3}x-1.3\) , therefore, less than sign is also used with “is equal to”.
Thus, the linear inequality is \(y\leqslant\frac{1}{3}x-1.3\) .
Thus, the linear equality represented by the graph is \(\boxed{y\leqslant\frac{1}{3}x-1.3}\) .
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Answer Details:
Grade: Junior High School
Subject: Mathematics
Chapter: Coordinate Geometry
Keywords: Coordinate Geometry, linear equation, system of linear equations in two variables, variables, mathematics, inequality.
