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Sagot :
In general, the equations of a line in point-slope form and slope-intercept form are:
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ y=mx+b \end{gathered}[/tex]respectively. Where m is the slope of the line, b is a constant, and (x_1, y_1) is a point on the line.
Thus, the point-slope form of the line described by the problem is:
[tex]y-3=m(x-5)[/tex]We simply need to calculate the slope of the line. For that, we simply require 2 points, we already have (5, 3) and, since the x-intercept is 6, we can deduce that the line goes through (0,6).
Therefore, the slope is:
[tex]m=\frac{y_2-y_1}{x_2-x_1}=\frac{6-3}{0-5}=-\frac{3}{5}[/tex]Then, the solution is:
[tex]y-3=-\frac{3}{5}(x-5)[/tex]
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