Answer :
To determine the distance between the points [tex]\((-4, -7)\)[/tex] and [tex]\((-8, -13)\)[/tex] on the coordinate plane, we can use the distance formula, which is derived from the Pythagorean Theorem. The distance formula is given by:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Here, the coordinates of the two points are:
- [tex]\((x_1, y_1) = (-4, -7)\)[/tex]
- [tex]\((x_2, y_2) = (-8, -13)\)[/tex]
First, let's calculate the differences in the x-coordinates ([tex]\(dx\)[/tex]) and y-coordinates ([tex]\(dy\)[/tex]):
[tex]\[ dx = x_2 - x_1 = -8 - (-4) = -8 + 4 = -4 \][/tex]
[tex]\[ dy = y_2 - y_1 = -13 - (-7) = -13 + 7 = -6 \][/tex]
Next, we square these differences:
[tex]\[ dx^2 = (-4)^2 = 16 \][/tex]
[tex]\[ dy^2 = (-6)^2 = 36 \][/tex]
We then add these squared differences together:
[tex]\[ dx^2 + dy^2 = 16 + 36 = 52 \][/tex]
Finally, we take the square root of this sum to find the distance:
[tex]\[ d = \sqrt{52} \][/tex]
Therefore, the distance between the points [tex]\((-4, -7)\)[/tex] and [tex]\((-8, -13)\)[/tex] is [tex]\(\sqrt{52}\)[/tex] units.
Thus, the correct answer is:
[tex]\[ \sqrt{52} \text{ units} \][/tex]
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Here, the coordinates of the two points are:
- [tex]\((x_1, y_1) = (-4, -7)\)[/tex]
- [tex]\((x_2, y_2) = (-8, -13)\)[/tex]
First, let's calculate the differences in the x-coordinates ([tex]\(dx\)[/tex]) and y-coordinates ([tex]\(dy\)[/tex]):
[tex]\[ dx = x_2 - x_1 = -8 - (-4) = -8 + 4 = -4 \][/tex]
[tex]\[ dy = y_2 - y_1 = -13 - (-7) = -13 + 7 = -6 \][/tex]
Next, we square these differences:
[tex]\[ dx^2 = (-4)^2 = 16 \][/tex]
[tex]\[ dy^2 = (-6)^2 = 36 \][/tex]
We then add these squared differences together:
[tex]\[ dx^2 + dy^2 = 16 + 36 = 52 \][/tex]
Finally, we take the square root of this sum to find the distance:
[tex]\[ d = \sqrt{52} \][/tex]
Therefore, the distance between the points [tex]\((-4, -7)\)[/tex] and [tex]\((-8, -13)\)[/tex] is [tex]\(\sqrt{52}\)[/tex] units.
Thus, the correct answer is:
[tex]\[ \sqrt{52} \text{ units} \][/tex]