Question
Point A
lies at (−4,4)
and point B
lies at (−3,6)
. Which point is needed so that △ABC
is scalene?
lies at (−4,4)
and point B
lies at (−3,6)
. Which point is needed so that △ABC
is scalene?
Answers
GPT-4o mini
To determine a point \( C \) that would make triangle \( ABC \) scalene, we first need to understand that a scalene triangle is one where all three sides have different lengths.
We have:
- Point \( A = (-4, 4) \)
- Point \( B = (-3, 6) \)
Let's first calculate the distance between points \( A \) and \( B \):
\[
d(A, B) = \sqrt{((-3) - (-4))^2 + (6 - 4)^2}
\]
\[
= \sqrt{(1)^2 + (2)^2}
\]
\[
= \sqrt{1 + 4} = \sqrt{5}
\]
Next, we need to find point \( C \) in such a way that the distances \( d(A, C) \), \( d(B, C) \), and \( d(A, B) \) are all different.
The distances between \( A \) and \( C \), and \( B \) and \( C \) can be expressed in terms of coordinates \( C = (x_C, y_C) \):
1. \( d(A, C) = \sqrt{(x_C + 4)^2 + (y_C - 4)^2} \)
2. \( d(B, C) = \sqrt{(x_C + 3)^2 + (y_C - 6)^2} \)
We can choose a point \( C \) that is neither on the line connecting \( A \) and \( B \) nor equidistant from \( A \) and \( B \), as that would create a situation where either one or both distances \( d(A, C) \) and \( d(B, C) \) might equal \( d(A, B) \), potentially forming an isosceles triangle.
For example, we can choose a point \( C \) that is clearly not on the line segment connecting \( A \) and \( B \). Let's try \( C = (0, 0) \):
Now, we calculate the distances using \( C(0, 0) \):
1. \( d(A, C) = \sqrt{(0 - (-4))^2 + (0 - 4)^2} = \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \)
2. \( d(B, C) = \sqrt{(0 - (-3))^2 + (0 - 6)^2} = \sqrt{(3)^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \)
Now we have the lengths:
- \( d(A, B) = \sqrt{5} \)
- \( d(A, C) = 4\sqrt{2} \)
- \( d(B, C) = 3\sqrt{5} \)
Since \( \sqrt{5} \), \( 4\sqrt{2} \), and \( 3\sqrt{5} \) are all different, triangle \( ABC \) is scalene.
Thus, the point \( C \) can be \( (0, 0) \) or any other point that doesn't result in two distances being equal. Other points can also be chosen, but any choice must ensure that the triangle remains scalene.
In conclusion, a point \( C = (0, 0) \) will ensure that the triangle \( ABC \) is scalene.
We have:
- Point \( A = (-4, 4) \)
- Point \( B = (-3, 6) \)
Let's first calculate the distance between points \( A \) and \( B \):
\[
d(A, B) = \sqrt{((-3) - (-4))^2 + (6 - 4)^2}
\]
\[
= \sqrt{(1)^2 + (2)^2}
\]
\[
= \sqrt{1 + 4} = \sqrt{5}
\]
Next, we need to find point \( C \) in such a way that the distances \( d(A, C) \), \( d(B, C) \), and \( d(A, B) \) are all different.
The distances between \( A \) and \( C \), and \( B \) and \( C \) can be expressed in terms of coordinates \( C = (x_C, y_C) \):
1. \( d(A, C) = \sqrt{(x_C + 4)^2 + (y_C - 4)^2} \)
2. \( d(B, C) = \sqrt{(x_C + 3)^2 + (y_C - 6)^2} \)
We can choose a point \( C \) that is neither on the line connecting \( A \) and \( B \) nor equidistant from \( A \) and \( B \), as that would create a situation where either one or both distances \( d(A, C) \) and \( d(B, C) \) might equal \( d(A, B) \), potentially forming an isosceles triangle.
For example, we can choose a point \( C \) that is clearly not on the line segment connecting \( A \) and \( B \). Let's try \( C = (0, 0) \):
Now, we calculate the distances using \( C(0, 0) \):
1. \( d(A, C) = \sqrt{(0 - (-4))^2 + (0 - 4)^2} = \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \)
2. \( d(B, C) = \sqrt{(0 - (-3))^2 + (0 - 6)^2} = \sqrt{(3)^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \)
Now we have the lengths:
- \( d(A, B) = \sqrt{5} \)
- \( d(A, C) = 4\sqrt{2} \)
- \( d(B, C) = 3\sqrt{5} \)
Since \( \sqrt{5} \), \( 4\sqrt{2} \), and \( 3\sqrt{5} \) are all different, triangle \( ABC \) is scalene.
Thus, the point \( C \) can be \( (0, 0) \) or any other point that doesn't result in two distances being equal. Other points can also be chosen, but any choice must ensure that the triangle remains scalene.
In conclusion, a point \( C = (0, 0) \) will ensure that the triangle \( ABC \) is scalene.