Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

[tex]\(\triangle ABC\)[/tex] is located at [tex]\(A(2,0)\)[/tex], [tex]\(B(4,4)\)[/tex], and [tex]\(C(6,3)\)[/tex]. Zackery says that [tex]\(\triangle ABC\)[/tex] is an isosceles triangle, while Verna says that it is a right triangle. Who is correct?

A. Zackery, because [tex]\(\overline{BC} \simeq \overline{AC}\)[/tex]

B. Zackery, because [tex]\(\overline{AB} \simeq \overline{BC}\)[/tex]

C. Verna, because [tex]\(\overline{BC} \perp \overline{AC}\)[/tex]

D. Verna, because [tex]\(\overline{AB} \perp \overline{BC}\)[/tex]


Sagot :

Let's find out whether [tex]$\triangle ABC$[/tex] is an isosceles triangle or a right triangle by computing the lengths of its sides and checking the necessary conditions for isosceles and right triangles.

First, let's determine the lengths of the sides [tex]\( \overline{AB} \)[/tex], [tex]\( \overline{BC} \)[/tex], and [tex]\( \overline{AC} \)[/tex].

### Length of [tex]\( \overline{AB} \)[/tex]
The points [tex]$A$[/tex] and [tex]$B$[/tex] are given as [tex]\( A(2,0) \)[/tex] and [tex]\( B(4,4) \)[/tex].
The distance formula is:
[tex]\[ AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} \][/tex]
Substituting the coordinates of [tex]$A$[/tex] and [tex]$B$[/tex]:
[tex]\[ AB = \sqrt{(4 - 2)^2 + (4 - 0)^2} = \sqrt{(2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.472 \][/tex]

### Length of [tex]\( \overline{BC} \)[/tex]
The points [tex]$B$[/tex] and [tex]$C$[/tex] are given as [tex]\( B(4,4) \)[/tex] and [tex]\( C(6,3) \)[/tex].
The distance formula is:
[tex]\[ BC = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2} \][/tex]
Substituting the coordinates of [tex]$B$[/tex] and [tex]$C$[/tex]:
[tex]\[ BC = \sqrt{(6 - 4)^2 + (3 - 4)^2} = \sqrt{(2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \approx 2.236 \][/tex]

### Length of [tex]\( \overline{AC} \)[/tex]
The points [tex]$A$[/tex] and [tex]$C$[/tex] are given as [tex]\( A(2,0) \)[/tex] and [tex]\( C(6,3) \)[/tex].
The distance formula is:
[tex]\[ AC = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2} \][/tex]
Substituting the coordinates of [tex]$A$[/tex] and [tex]$C$[/tex]:
[tex]\[ AC = \sqrt{(6 - 2)^2 + (3 - 0)^2} = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \][/tex]

### Checking for Isosceles Triangle
To check if [tex]$\triangle ABC$[/tex] is isosceles, we need to see if any two sides are of equal length.

We have:
[tex]\[ AB \approx 4.472, \quad BC \approx 2.236, \quad AC = 5 \][/tex]
Since none of the lengths are approximately equal, [tex]\(\triangle ABC\)[/tex] is not isosceles.

### Checking for Right Triangle
To check if [tex]$\triangle ABC$[/tex] is a right triangle, we need to use the Pythagorean theorem and verify if any combination of sides satisfies:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
Let's test the following combinations:

1. [tex]\(AB^2 + BC^2 \approx AC^2\)[/tex]:
[tex]\[ (4.472)^2 + (2.236)^2 \approx 5^2 \][/tex]
[tex]\[ 20 + 5 \approx 25 \quad \text{(True)} \][/tex]

Since [tex]\(AB^2 + BC^2 = AC^2\)[/tex], we confirm that [tex]\(\triangle ABC\)[/tex] is a right triangle.

Therefore, Verna is correct because [tex]\(\overline{BC} \perp \overline{AC}\)[/tex].

Answer: Verna, because [tex]\(\overline{BC} \perp \overline{AC}\)[/tex]
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.