Question

Each set of coordinate pairs identifies the vertices of a right triangle. Select all sets that identify right triangles with areas equal to 10 square units.
A. {(0, 1), (1, 4), (–6, 3)}
B. {(3, 0), (–3, 2), (4, 3)}
C. {(–1, –2), (–4, –1), (–3, 2)}
D. {(–1, –2), (–2, 5), (–4, –1)}
E. {(–2, –5), (–5, –4), (–3, 2)}

Answer 1

The area of a right triangle is given by half the product of the leg lengths.

The vertices of the right triangles with areas equal to 10 square units are;

• A. {(0, 1), (1, 4), (-6, 3)}

• B. {(3, 0), (-3, 2), (4, 3)}

• D. {(-1, -2), (-2, 5), (-4, -1)}

• E. {(-2, -5), (-5, -4), (-3, 2)}

Reasons:

The slope of the legs of a right triangle are the negative inverse reciprocal of the other.

The slope of a line with points (x₁, y₁), and (x₂, y₂) is given by the equation;

$Slope, \, m = \mathbf{\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}}$

Length of side, $l = \mathbf{ \sqrt{\left (x_{2}-x_{1} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}}$

Area of a right triangle, A = 0.5 × Product of the length of the legs

Option A

In option A, we have; {(0, 1), (1, 4), (-6, 3)}

$\displaystyle Slope \ of \ side \ a = \mathbf{ \frac{4 - 1}{1 - 0}} = 3$

$\displaystyle Slope \ of \ side \ b = \frac{4 - 3}{1 - (-6)} = \frac{1}{7}$

$\displaystyle Slope \ of \ side \ c = \frac{3 - 1}{(-6) - 0} = -\frac{2}{6} =-\frac{1}{3}$

Therefore, option a is a right triangle, with side a perpendicular to side c

Which gives;

Length of side a = √((1 - 0)² + (4 - 1)²) = √(10)

Length of side c = √(((-6) - 0)² + (3 - 1)²) = √(40) = 2·√(10)

Area of triangle in option A, A = 0.5 × √(10) × 2·√(10) = 10

Therefore;

• The right triangle with vertices {(0, 1), (1, 4), (-6, 3)} has an area of 10 square units.

Option B.

In option B, we have; {(3, 0), (-3, 2), (4, 3)}

$\displaystyle Slope \ of \ side \ a = \mathbf{\frac{2 - 0}{(-3) - 3}} = -\frac{2}{6} = -\frac{1}{3}$

$\displaystyle Slope \ of \ side \ b = \frac{3- 2}{4 - (-3)} = \frac{1}{7}$

$\displaystyle Slope \ of \ side \ c = \frac{3 - 0}{4 - 3} = 3$

Therefore, option B is a right triangle, with side a perpendicular to side c

Which gives;

Length of side a = √((2 - 0)² + ((-3) - 3)²) = √(2² + (-6)²) = √(40) = 2·√10

Length of side c = √((4 - 3)² + (3 - 0)²) = √(1² + 3²) = √(10)

Area, A = 2·√(10) × √(10) = 10

Therefore;

• The area of the right triangle with vertices {(3, 0), (-3, 2), (4, 3)} is 10 square units.

Option C.

In option C, we have; {(-1, -2), (-4, -1), (-3, 2)}

$\displaystyle Slope \ of \ side \ a = \mathbf{\frac{-4 - (-1)}{(-1) - (-2)} = -\frac{3}{1}} = -3$

$\displaystyle Slope \ of \ side \ b = \frac{2- (-1)}{-3 - (-4)} =3$

$\displaystyle Slope \ of \ side \ c = \frac{2 - (-2)}{(-3) - (-1)} = -2$

Therefore, option C is not a right triangle

Option D.

In option D, we have; {(-1, -2), (-2, 5), (-4, -1)}

$\displaystyle Slope \ of \ side \ a = \mathbf{\frac{5 - (-2)}{(-2) - (-1)}} = -\frac{7}{1} = -7$

$\displaystyle Slope \ of \ side \ b = \frac{5- (-1)}{(-2) - (-4)} = \frac{6}{2} = 3$

$\displaystyle Slope \ of \ side \ c = \frac{(-1) - (-2)}{(-4) - (-1)} = -\frac{1}{3}$

Therefore;

Option D is a right triangle, with side b perpendicular to side c

Which gives;

Length of side b = √(((-2) - (-4))² + (5 - (-1))²) = √(2² + (-6)²) = √(40) = 2·√(10)

Length of side c = √(((-4) - (-1))² + ((-1) - (-2))²) = √((-3)² + 1²) = √(10)

Area of the right triangle in option D, A = 0.5 × 2·√(10) × √(10) = 10

Therefore;

• The right triangle with vertices {(-1, -2), (-2, 5), (-4, -1)}, has an area of 10 square units.

Option E.

In option E, we have; {(-2, -5), (-5, -4), (-3, 2)}

$\displaystyle \mathbf{Slope \ of \ side} \ a = \frac{(-4) - (-5)}{(-5) - (-2)} = -\frac{1}{3}$

$\displaystyle Slope \ of \ side \ b = \frac{-4- 2}{(-5) - (-3)} = \frac{6}{2} = 3$

Therefore, option E is a right triangle, with side a perpendicular to side b

Which gives;

Length of side a = √(((-5) - (-2))² +((-4) - (-5))²) = √((-3)² + 1²) = √(10)

Length of side b = √(((-5) - (-3))² + (-4 - 2)²) = √((-2)² + (-6)²) = √(40) = 2·√(10)

The area of right triangle, A = 0.5 × √(10) × 2·√(10) = 10

• The right triangle with vertices {(-2, -5), (-5, -4), (-3, 2)} has an area equal to 10 square units.

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