Answer
See graph
Step-by-step explanation
The given line segment passes through the point [tex](-5,7)[/tex] and [tex](4,-3)[/tex].
We plot this points and draw a straight line to them to obtain the blue line as shown in the
attachment.
The slope of this line segment is given by the formula
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
This implies that,
[tex]m=\frac{7--3}{-5-4}[/tex]
[tex]m=-\frac{10}{9}[/tex]
The equation of this line is given by the formula,
[tex]y-y_1=m(x-x_1)[/tex]
[tex]y+3=-\frac{10}{9}(x-4)[/tex]
[tex]9y+27=-10(x-4)[/tex]
[tex]9y+27=-10x+40[/tex]
[tex]9y=-10x+13[/tex]
We now find the inverse by interchanging x and y.
[tex]9x=-10y+13[/tex]
We now make y the subject,
[tex]9x-13=-10y[/tex]
[tex]y=-\frac{9}{10}x+\frac{13}{10}[/tex]
We now plot some points.
When
[tex]x=7[/tex]
[tex]y=-\frac{9}{10}\times 7+\frac{13}{10}[/tex]
[tex]y=-\frac{50}{10}=-5[/tex]
This gives the ordered pair
[tex](7,-5)[/tex]
Also when
[tex]x=-3[/tex]
[tex]y=-\frac{9}{10}\times -3+\frac{13}{10}[/tex]
[tex]y=\frac{40}{10}=4[/tex]
This gives the ordered pairs,
[tex](-3,4)[/tex]
We plot these points too and draw a straight line through them to get the red line in the attachment.
