Question

Plot each point and form the triangle \(A B C\). Show that the triangle is a right triangle. Find its area. $$ A=(-5,3) ; \quad B=(6,0) ; \quad C=(5,5) $$

Short answer

The triangle ABC is a right triangle and its area is 26 square units.

Step-by-step solution

- Plot Points on Coordinate Plane

Plot the points A(-5,3), B(6,0), and C(5,5) on the coordinate plane. Label each point accordingly.

- Determine Distances Between Points

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) to find the lengths of sides AB, BC, and AC.\(AB = \sqrt{(6 - (-5))^2 + (0 - 3)^2} = \sqrt{11^2 + (-3)^2} = \sqrt{121 + 9} = \sqrt{130}\)\(BC = \sqrt{(6 - 5)^2 + (0 - 5)^2} = \sqrt{1^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26}\)\(AC = \sqrt{(5 - (-5))^2 + (5 - 3)^2} = \sqrt{10^2 + 2^2} = \sqrt{100 + 4} = \sqrt{104}\)

- Verify Right Triangle Using Pythagorean Theorem

Check if the triangle is a right triangle by verifying if the sum of the squares of two sides equals the square of the longest side.\(AB^2 + BC^2 = (\sqrt{130})^2 + (\sqrt{26})^2 = 130 + 26 = 156\)\(AC^2 = (\sqrt{104})^2 = 104\)Since \(AB^2 + AC^2 eq BC^2\), check another pair:\(AC^2 + BC^2 = (\sqrt{104})^2 + (\sqrt{26})^2 = 104 + 26 = 130\)\(AB^2 = (\sqrt{130})^2 = 130\)This confirms that \(AC^2 + BC^2 = AB^2\), hence ABC is a right triangle.

- Calculate the Area of the Triangle

The area of a right triangle can be calculated by using the formula \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\). Here, the base and the height can be AC and BC respectively.\[\text{Area} = \frac{1}{2} \times AC \times BC = \frac{1}{2} \times \sqrt{104} \times \sqrt{26} = \frac{1}{2} \times \sqrt{2704} = \frac{1}{2} \times 52 = 26\]

Key concepts

coordinate geometry

In coordinate geometry, we use a plane defined by two perpendicular lines called axes. Each point on this plane is represented by a pair of numbers, known as coordinates, which tell us its exact location.

• The horizontal axis is called the x-axis.
• The vertical axis is called the y-axis.

For example, the point A(-5, 3) means that point A is located at -5 on the x-axis and 3 on the y-axis. By plotting the given points (A, B, and C) on this plane, we can visualize the shape of the triangle formed.

distance formula

The distance formula helps us find the length between two points in a coordinate plane. It is derived from the Pythagorean theorem.

• The formula is: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

For example, to find the distance between points A(-5, 3) and B(6, 0):
\[AB = \sqrt{(6 - (-5))^2 + (0 - 3)^2} = \sqrt{11^2 + (-3)^2} = \sqrt{121 + 9} = \sqrt{130} \]
Repeating similar calculations for other pairs gives \(BC = \sqrt{26} \) and \(AC = \sqrt{104}\). This distance calculation is crucial for determining the triangle's properties, such as its type and area.

Pythagorean theorem

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

• The formula is: \(c^2 = a^2 + b^2\)

For our triangle ABC, first, we check:\
\(AB^2 + AC^2 = 130 + 26 = 156\)
\(BC^2 = 104\)
Since this doesn't match, we test another pair:
\[AC^2 + BC^2 = 104 + 26 = 130 \]
\[AB^2 = 130 \]
Thus, \(AC^2 + BC^2 = AB^2\), proving that triangle ABC is indeed a right triangle.

triangle area calculation

Calculating the area of a right triangle is straightforward when its base and height are known. The formula to find the area of a triangle is:

• \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)

Here, we can take AC and BC as the base and height respectively:
\[ \text{Area} = \frac{1}{2} \times \text{AC} \times \text{BC} \]
We know \(AC = \frac{1}{2} \times \text{AC} \times \text{BC} = \frac{1}{2} \times 52 \)
This simplifies to \ \text{Area} = 26 \
So, the area of triangle ABC is 26 square units.