The graph of an equation is shown below:line joining ordered pairs negative 3, negative 2 and 1, 3Based on the graph, which of the following represents a solution to the equation? (−2, −3) (3, 1) (1, 3) (3, 2)

Answer:The point   [tex](1,3)[/tex] is a solution of the linear equationStep-by-step explanation:we have[tex]A(-3,-2)\ B(1,3)[/tex]Find the equation of the line ABThe slope of the line is equal to[tex]m=\frac{y2-y1}{x2-x1}[/tex]substitute the values[tex]m=\frac{3+2}{1+3}[/tex][tex]m=\frac{5}{4}[/tex]Find the equation of the into point-slope form[tex]y-y1=m(x-x1)[/tex]we have[tex]m=\frac{5}{4}[/tex][tex](x1,y1)=B(1,3)[/tex]substitute[tex]y-3=\frac{5}{4}(x-1)[/tex]we know thatIf a point is a solution to the equation of the linethen the point must be satisfy the equation and based on the graph the point must be on the lineLet's check every point.case A) point [tex](-2,-3)[/tex]Substitute the value of x and y in the equation of the line[tex]-3-3=\frac{5}{4}(-2-1)[/tex][tex]-6=-\frac{15}{4}[/tex] -------> is not trueThe point  [tex](-2,-3)[/tex] is not a solution of the linear equationSee the attached figure-------> the point is not on the linecase B) point [tex](3,1)[/tex]Substitute the value of x and y in the equation of the line[tex]1-3=\frac{5}{4}(3-1)[/tex][tex]-2=\frac{10}{4}[/tex] -------> is not trueThe point   [tex](3,1)[/tex] is not a solution of the linear equationSee the attached figure-------> the point is not on the linecase C) point [tex](1,3)[/tex]Substitute the value of x and y in the equation of the line[tex]3-3=\frac{5}{4}(1-1)[/tex][tex]0=0[/tex] -------> is  trueThe point   [tex](1,3)[/tex] is a solution of the linear equationSee the attached figure-------> the point is on the linecase D) point [tex](3,2)[/tex]Substitute the value of x and y in the equation of the line[tex]2-3=\frac{5}{4}(3-1)[/tex][tex]-1=\frac{10}{4}[/tex] -------> is not trueThe point   [tex](3,2)[/tex] is not a solution of the linear equationSee the attached figure-------> the point is not on the line