The area of a triangle with vertices at coordinates \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) can be found using the formula:
\(\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|\)
Given the vertices \( A(2,3) \), \( B(7,1) \), and \( C(4,6) \), we can substitute these into the formula as follows:
\(\text{Area} = \frac{1}{2} \left| 2(1-6) + 7(6-3) + 4(3-1) \right|\)
Simplifying each term gives:
- \(2(1-6) = 2 \times -5 = -10\)
- \(7(6-3) = 7 \times 3 = 21\)
- \(4(3-1) = 4 \times 2 = 8\)
Substitute back into the formula and calculate:
\(\text{Area} = \frac{1}{2} \left| -10 + 21 + 8 \right| = \frac{1}{2} \left| 19 \right|\)
\(\text{Area} = \frac{1}{2} \times 19 = 9.5\) square units.
Therefore, the area of the triangle is 9.5 square units.
Similar Questions
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Important Questions from Coordinate Geometry
What is the reflection of the point (-1, 5) in the line x = 1?
What are the co-ordinates of the centroid of a triangle, whose vertices are A(1, -5), B(-4, 0) and C(3, -4)?
Slope of the line AB is 4/3. Co-ordinates of points A and B are (x, -5) and (2, -3) respectively. What is the value of x?
Find the co-ordinates of the centroid of a triangle whose vertices are A(1, 4), B(7, 8) and C(10, 12).
- If x² + y² - 12x + 18y + 117 = 0, then the value of x² + y² is:
