Question

If the vertices of triangle are (-2,6), (3,-6) and (1,5), then the area of the triangle is

• A. 40 sq. units
• B. 30 sq. units
• C. 15.5 sq. units
• D. 35 sq. units

Answer 1

The Correct Option is C

Let's label the vertices of the triangle as \(A(-2,6), B(3,-6), and\ C(1,5).\)
The formula to calculate the area of a triangle formed by three points \(A(x_1, y_1), B(x_2, y_2), and\  C(x_3, y_3)\) is:
\(\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|\)
Substituting the coordinates of the vertices into the formula:
\(\text{Area} = \frac{1}{2} \left| -2(-6 - 5) + 3(5 - (-6)) + 1(6 - (-6)) \right|\)
 \(=\frac{1}{2} \left| -2(-11) + 3(-1) + 1(12) \right|\)
\(=\) \(\frac{1}{2} \times (22 - 3 + 12)\)
= \(\frac{1}{2} \times 31\)
= \(\frac{31}{2}\)
= 15.5
Therefore, the area of the triangle is 15.5 square units (option C).

Answer 2

Let the vertices of the triangle be: \( A(-2,6), B(3,-6), C(1,5) \) The area of a triangle with vertices \( (x_1,y_1), (x_2,y_2), (x_3,y_3) \) is given by: \[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] Substitute the coordinates: \[ = \frac{1}{2} \left| -2(-6 - 5) + 3(5 - 6) + 1(6 - (-6)) \right| \] \[ = \frac{1}{2} \left| -2(-11) + 3(-1) + 1(12) \right| \] \[ = \frac{1}{2} \left| 22 - 3 + 12 \right| = \frac{1}{2} \cdot 31 = \mathbf{15.5 \text{ sq. units}} \]

Answer 3

Let the vertices of the triangle be A(-2, 6), B(3, -6), and C(1, 5).

The area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is given by the formula:

Area = \(\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\)

Plugging in the coordinates of A, B, and C:

Area = \(\frac{1}{2} |(-2)(-6 - 5) + (3)(5 - 6) + (1)(6 - (-6))|\)

Area = \(\frac{1}{2} |(-2)(-11) + (3)(-1) + (1)(12)|\)

Area = \(\frac{1}{2} |22 - 3 + 12|\)

Area = \(\frac{1}{2} |31|\)

Area = \(\frac{31}{2}\)

Area = 15.5 sq. units

Answer: 15.5 sq. units