Answer:
The point [tex](1,3)[/tex] is a solution of the linear equation
Step-by-step explanation:
we have
[tex]A(-3,-2)\ B(1,3)[/tex]
Find the equation of the line AB
The 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 that
If a point is a solution to the equation of the line
then
the point must be satisfy the equation and based on the graph the point must be on the line
Let'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 true
The point [tex](-2,-3)[/tex] is not a solution of the linear equation
See the attached figure-------> the point is not on the line
case 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 true
The point [tex](3,1)[/tex] is not a solution of the linear equation
See the attached figure-------> the point is not on the line
case 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 true
The point [tex](1,3)[/tex] is a solution of the linear equation
See the attached figure-------> the point is on the line
case 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 true
The point [tex](3,2)[/tex] is not a solution of the linear equation
See the attached figure-------> the point is not on the line
