To prove that the point [tex]\((4,4)\)[/tex] lies on the locus of points equidistant from [tex]\((1,3)\)[/tex] and [tex]\((5,7)\)[/tex], we need to show that the distance from [tex]\((4,4)\)[/tex] to [tex]\((1,3)\)[/tex] is the same as the distance from [tex]\((4,4)\)[/tex] to [tex]\((5,7)\)[/tex].
1. Distance Formula: The formula to calculate the distance between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\][/tex]
2. Calculate Distance from [tex]\((4,4)\)[/tex] to [tex]\((1,3)\)[/tex]:
[tex]\[
d_1 = \sqrt{(4 - 1)^2 + (4 - 3)^2}
\][/tex]
Simplifying inside the square root:
[tex]\[
d_1 = \sqrt{(3)^2 + (1)^2}
\][/tex]
[tex]\[
d_1 = \sqrt{9 + 1}
\][/tex]
[tex]\[
d_1 = \sqrt{10}
\][/tex]
Therefore,
[tex]\[
d_1 \approx 3.1622776601683795
\][/tex]
3. Calculate Distance from [tex]\((4,4)\)[/tex] to [tex]\((5,7)\)[/tex]:
[tex]\[
d_2 = \sqrt{(4 - 5)^2 + (4 - 7)^2}
\][/tex]
Simplifying inside the square root:
[tex]\[
d_2 = \sqrt{(-1)^2 + (-3)^2}
\][/tex]
[tex]\[
d_2 = \sqrt{1 + 9}
\][/tex]
[tex]\[
d_2 = \sqrt{10}
\][/tex]
Therefore,
[tex]\[
d_2 \approx 3.1622776601683795
\][/tex]
4. Compare the Distances:
[tex]\[
d_1 \approx 3.1622776601683795
\][/tex]
[tex]\[
d_2 \approx 3.1622776601683795
\][/tex]
Since [tex]\(d_1\)[/tex] is equal to [tex]\(d_2\)[/tex], the point [tex]\((4,4)\)[/tex] is equidistant from both [tex]\((1,3)\)[/tex] and [tex]\((5,7)\)[/tex].
Therefore, the point [tex]\((4,4)\)[/tex] lies on the locus of points which are equidistant from [tex]\((1,3)\)[/tex] and [tex]\((5,7)\)[/tex].
