Get detailed and accurate responses to your questions with IDNLearn.com. Ask your questions and receive comprehensive and trustworthy answers from our experienced community of professionals.
Sagot :
To determine whether [tex]\( \triangle ABC \)[/tex] is an isosceles triangle, a right triangle, or both, we need to analyze the coordinates of the vertices of the triangle. The coordinates are:
[tex]\[ A = (2, 0), \quad B = (4, 4), \quad C = (6, 3) \][/tex]
Step 1: Calculate the lengths of the sides using the distance formula.
The distance formula for two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Calculating [tex]\( AB \)[/tex]:
[tex]\[ AB = \sqrt{(4-2)^2 + (4-0)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \][/tex]
Calculating [tex]\( BC \)[/tex]:
[tex]\[ BC = \sqrt{(6-4)^2 + (3-4)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \][/tex]
Calculating [tex]\( AC \)[/tex]:
[tex]\[ AC = \sqrt{(6-2)^2 + (3-0)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \][/tex]
Step 2: Determine if the triangle is isosceles by checking if any two sides are equal.
[tex]\[ AB = 2\sqrt{5}, \quad BC = \sqrt{5}, \quad AC = 5 \][/tex]
No two sides are equal. Therefore, [tex]\( \triangle ABC \)[/tex] is not an isosceles triangle.
Step 3: Determine if the triangle is a right triangle by checking the dot product of vectors.
First, define the vectors:
[tex]\[ \overrightarrow{AB} = (B_x - A_x, B_y - A_y) = (4 - 2, 4 - 0) = (2, 4) \][/tex]
[tex]\[ \overrightarrow{BC} = (C_x - B_x, C_y - B_y) = (6 - 4, 3 - 4) = (2, -1) \][/tex]
[tex]\[ \overrightarrow{AC} = (C_x - A_x, C_y - A_y) = (6 - 2, 3 - 0) = (4, 3) \][/tex]
Calculate the dot products:
[tex]\[ \overrightarrow{AB} \cdot \overrightarrow{BC} = (2, 4) \cdot (2, -1) = (2)(2) + (4)(-1) = 4 - 4 = 0 \][/tex]
Since the dot product [tex]\(\overrightarrow{AB} \cdot \overrightarrow{BC} = 0\)[/tex], the vectors [tex]\(\overrightarrow{AB}\)[/tex] and [tex]\(\overrightarrow{BC}\)[/tex] are perpendicular. Therefore, [tex]\( \triangle ABC \)[/tex] is a right triangle with [tex]\( \overline{AB} \)[/tex] perpendicular to [tex]\( \overline{BC} \)[/tex].
Conclusion:
Since [tex]\( \triangle ABC \)[/tex] is a right triangle and not an isosceles triangle, Verna is correct.
Therefore, the answer is:
[tex]\[ \boxed{4} \quad \text{Verna, because} \ \overline{AB} \perp \overline{BC} \][/tex]
[tex]\[ A = (2, 0), \quad B = (4, 4), \quad C = (6, 3) \][/tex]
Step 1: Calculate the lengths of the sides using the distance formula.
The distance formula for two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Calculating [tex]\( AB \)[/tex]:
[tex]\[ AB = \sqrt{(4-2)^2 + (4-0)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \][/tex]
Calculating [tex]\( BC \)[/tex]:
[tex]\[ BC = \sqrt{(6-4)^2 + (3-4)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \][/tex]
Calculating [tex]\( AC \)[/tex]:
[tex]\[ AC = \sqrt{(6-2)^2 + (3-0)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \][/tex]
Step 2: Determine if the triangle is isosceles by checking if any two sides are equal.
[tex]\[ AB = 2\sqrt{5}, \quad BC = \sqrt{5}, \quad AC = 5 \][/tex]
No two sides are equal. Therefore, [tex]\( \triangle ABC \)[/tex] is not an isosceles triangle.
Step 3: Determine if the triangle is a right triangle by checking the dot product of vectors.
First, define the vectors:
[tex]\[ \overrightarrow{AB} = (B_x - A_x, B_y - A_y) = (4 - 2, 4 - 0) = (2, 4) \][/tex]
[tex]\[ \overrightarrow{BC} = (C_x - B_x, C_y - B_y) = (6 - 4, 3 - 4) = (2, -1) \][/tex]
[tex]\[ \overrightarrow{AC} = (C_x - A_x, C_y - A_y) = (6 - 2, 3 - 0) = (4, 3) \][/tex]
Calculate the dot products:
[tex]\[ \overrightarrow{AB} \cdot \overrightarrow{BC} = (2, 4) \cdot (2, -1) = (2)(2) + (4)(-1) = 4 - 4 = 0 \][/tex]
Since the dot product [tex]\(\overrightarrow{AB} \cdot \overrightarrow{BC} = 0\)[/tex], the vectors [tex]\(\overrightarrow{AB}\)[/tex] and [tex]\(\overrightarrow{BC}\)[/tex] are perpendicular. Therefore, [tex]\( \triangle ABC \)[/tex] is a right triangle with [tex]\( \overline{AB} \)[/tex] perpendicular to [tex]\( \overline{BC} \)[/tex].
Conclusion:
Since [tex]\( \triangle ABC \)[/tex] is a right triangle and not an isosceles triangle, Verna is correct.
Therefore, the answer is:
[tex]\[ \boxed{4} \quad \text{Verna, because} \ \overline{AB} \perp \overline{BC} \][/tex]
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Your questions deserve precise answers. Thank you for visiting IDNLearn.com, and see you again soon for more helpful information.