Certainly! Let's proceed step-by-step to demonstrate that for points [tex]\( A(6, -1) \)[/tex], [tex]\( B(1, 3) \)[/tex], and [tex]\( C(-2, 9) \)[/tex], the distance [tex]\( AC \)[/tex] is indeed twice the distance [tex]\( AB \)[/tex].
1. Calculate the distance [tex]\( AB \)[/tex]:
The distance formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\][/tex]
For points [tex]\( A(6, -1) \)[/tex] and [tex]\( B(1, 3) \)[/tex]:
[tex]\[
AB = \sqrt{(1 - 6)^2 + (3 + 1)^2} = \sqrt{(-5)^2 + (4)^2} = \sqrt{25 + 16} = \sqrt{41}
\][/tex]
Approximating the value:
[tex]\[
AB \approx 6.4031242374328485
\][/tex]
2. Calculate the distance [tex]\( AC \)[/tex]:
For points [tex]\( A(6, -1) \)[/tex] and [tex]\( C(-2, 9) \)[/tex]:
[tex]\[
AC = \sqrt{(-2 - 6)^2 + (9 + 1)^2} = \sqrt{(-8)^2 + (10)^2} = \sqrt{64 + 100} = \sqrt{164}
\][/tex]
Approximating the value:
[tex]\[
AC \approx 12.806248474865697
\][/tex]
3. Verify if [tex]\( AC = 2 AB \)[/tex]:
We need to check if:
[tex]\[
\sqrt{164} = 2 \cdot \sqrt{41}
\][/tex]
Given the calculated numerical approximations:
[tex]\[
12.806248474865697 \approx 2 \times 6.4031242374328485
\][/tex]
Hence, we confirm that:
[tex]\[
AC = 2 AB
\][/tex]
So, we have shown that [tex]\( AC = 2 AB \)[/tex] for points [tex]\( A(6, -1) \)[/tex], [tex]\( B(1, 3) \)[/tex], and [tex]\( C(-2, 9) \)[/tex].
