The slope of line XZ is the negative reciprocal of the slope of line CA because the triangles are flipped.

The slope of line XZ is one-half the slope of line CA because the ratio of corresponding sides between the triangles is  $\frac{1}{2}$  .

The slope of line XZ is the same as the slope of line CA because the ratios of the corresponding sides of the triangles are equivalent.

The is no relationship between the slope of line XZ and the slope of line CA.

$\frac{1-\left(-5\right)}{1-\left(-2\right)}=\frac{1-3}{1-\left(-5\right)}$

$\frac{1-\left(-5\right)}{1-\left(-2\right)}=\frac{5-1}{3-1}$

$\frac{-2-\left(-5\right)}{-2-1}=\frac{5-1}{3-1}$

$\frac{1-\left(-2\right)}{1-\left(-5\right)}=\frac{3-1}{5-1}$

The slope of line HK is less than the slope of line PK because the ratio of the change in y-values of the endpoints to the change in x-values of the endpoints is less for line HK than it is for line PK.

The slope of line HK is equal to the slope of line PK because the ratio of the change in y-values of the endpoints to the change in x-values of the endpoints is the same for line HK as it is for line PK.

The slope of line HK is greater than the slope of line PK because the ratio of the change in y-values of the endpoints to the change in x-values of the endpoints is greater for line HK than it is for line PK.

The relationship between the slope of line HK and the slope of line PK cannot be determined because the traingles are congruent.

$\frac{3-7}{-4-\left(-7\right)}=\frac{-5-3}{2-\left(-4\right)}$

$\frac{3-\left(-4\right)}{7-\left(-7\right)}=\frac{-5-2}{3-\left(-4\right)}$

$\frac{-4-\left(-7\right)}{3-7}=\frac{2-\left(-4\right)}{-5-3}$

$\frac{-4-\left(-3\right)}{-7-7}=\frac{2-\left(-5\right)}{-4-3}$

2

1/2

-2

-1/2

1

-1

3

-3