Select the correct answer.

Two points located on [tex]$\stackrel{\rightharpoonup}{ JK }$[/tex] are [tex]$J(-1,-9)$[/tex] and [tex]$K(5,3)$[/tex]. What is the slope of [tex]$\stackrel{\rightharpoonup}{ JK }$[/tex]?

A. -2
B. [tex]$-\frac{1}{2}$[/tex]
C. [tex]$\frac{1}{2}$[/tex]
D. 2



Answer :

To find the slope of the line that passes through two points [tex]\( J(-1, -9) \)[/tex] and [tex]\( K(5, 3) \)[/tex], you can use the slope formula:
[tex]\[ \text{slope} = \frac{{y_2 - y_1}}{{x_2 - x_1}} \][/tex]

Here, the coordinates of point [tex]\( J \)[/tex] ([tex]\( x_1, y_1 \)[/tex]) are [tex]\(-1, -9\)[/tex] and the coordinates of point [tex]\( K \)[/tex] ([tex]\( x_2, y_2 \)[/tex]) are [tex]\(5, 3\)[/tex].

Substitute the given coordinates into the formula:

[tex]\[ \text{slope} = \frac{{y_2 - y_1}}{{x_2 - x_1}} = \frac{3 - (-9)}{5 - (-1)} \][/tex]

Simplify inside the numerator and the denominator:

[tex]\[ \text{slope} = \frac{3 + 9}{5 + 1} = \frac{12}{6} \][/tex]

Finally, divide the numerator by the denominator to find the slope:

[tex]\[ \text{slope} = \frac{12}{6} = 2 \][/tex]

Therefore, the slope of [tex]\( \stackrel{\rightharpoonup}{JK} \)[/tex] is [tex]\( 2 \)[/tex].

The correct answer is:
D. [tex]\( 2 \)[/tex]