Question

The points A(-5,0), B(5,0) and C(0,4) are the vertices of which triangle?

• A. A right-angled triangle
• B. An equilateral triangle
• C. An isosceles triangle
• D. A scalene triangle

Answer 1

The Correct Option is C

Given vertices:

\( A(-5,0) \), \( B(5,0) \), and \( C(0,4) \) 

Step 1: Calculate the side lengths

Using the distance formula:

\( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

AB:

\( AB = \sqrt{(5 - (-5))^2 + (0 - 0)^2} = \sqrt{(5+5)^2} = \sqrt{10^2} = 10 \)

BC:

\( BC = \sqrt{(0 - 5)^2 + (4 - 0)^2} = \sqrt{(-5)^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41} \)

AC:

\( AC = \sqrt{(0 - (-5))^2 + (4 - 0)^2} = \sqrt{5^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41} \)

Step 2: Identify the triangle type

- Since \( AC = BC \), the triangle is isosceles.

- To check if it is right-angled, we use the Pythagorean theorem:

\( AB^2 = AC^2 + BC^2 \)

\( 10^2 = (\sqrt{41})^2 + (\sqrt{41})^2 \)

\( 100 = 41 + 41 = 82 \), which is not true.

So, the triangle is isosceles but not right-angled.

Final Answer: An isosceles triangle

Answer 2

An isosceles triangle has two sides of equal length. To determine if the triangle with vertices A(-5,0), B(5,0), and C(0,4) is isosceles, we calculate the distances between each pair of points using the distance formula:

d = √((x2-x1)² + (y2-y1)²)

Step 1: Calculate AB
Coordinates: A(-5,0), B(5,0)
d_AB= √((5 - (-5))² + (0 - 0)²) = √((10)² + 0) = √100 = 10

Step 2: Calculate AC
Coordinates: A(-5,0), C(0,4)
d_AC= √((0 - (-5))² + (4 - 0)²) = √(5² + 4²) = √(25 + 16) = √41

Step 3: Calculate BC
Coordinates: B(5,0), C(0,4)
d_BC= √((0 - 5)² + (4 - 0)²) = √(5² + 4²) = √(25 + 16) = √41

The distances AB = 10, AC = √41, and BC = √41 show that two sides (AC and BC) are equal, indicating that triangle ABC is an isosceles triangle.