Question

a line segment has end points (-5, 7) and (4,-3) find the inverse of the line segment and graph both the segment and its inverse.

Answer 1

Answer

See graph

Step-by-step explanation

The given line segment passes through the point $(-5,7)$ and  $(4,-3)$.

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

$m=\frac{y_2-y_1}{x_2-x_1}$

This implies that,

$m=\frac{7--3}{-5-4}$

$m=-\frac{10}{9}$

The equation of this line is given by the formula,

$y-y_1=m(x-x_1)$

$y+3=-\frac{10}{9}(x-4)$

$9y+27=-10(x-4)$

$9y+27=-10x+40$

$9y=-10x+13$

We now find the inverse by interchanging x and y.

$9x=-10y+13$

We now make y the subject,

$9x-13=-10y$

$y=-\frac{9}{10}x+\frac{13}{10}$

We now plot some points.

When

$x=7$

$y=-\frac{9}{10}\times 7+\frac{13}{10}$

$y=-\frac{50}{10}=-5$

This gives the ordered pair

$(7,-5)$

Also when

$x=-3$

$y=-\frac{9}{10}\times -3+\frac{13}{10}$

$y=\frac{40}{10}=4$

This gives the ordered pairs,

$(-3,4)$

We plot these points too and draw a straight line through them to get the red line in the attachment.