Question

Area of Triangle with point (1,2) (3,6) (5,10) is

• A. 0
• B. 2
• C. 4
• D. 6

Hint:

Before calculating the full area, check if the points follow a linear pattern. Here, for every increase of 2 in $x$, $y$ increases by 4 ($y = 2x$). Since all points satisfy this, they are collinear and the area is 0.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The area of a triangle formed by three points can be calculated using their coordinates. If the calculated area is zero, it implies that the three points do not form a triangle because they lie on a single straight line (collinear).

Step 2: Key Formula or Approach:
For points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, the area is: \[ \text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \]

Step 3: Detailed Explanation:
Given points: $(1, 2), (3, 6), (5, 10)$. \[ \text{Area} = \frac{1}{2} |1(6 - 10) + 3(10 - 2) + 5(2 - 6)| \] \[ \text{Area} = \frac{1}{2} |1(-4) + 3(8) + 5(-4)| \] \[ \text{Area} = \frac{1}{2} |-4 + 24 - 20| \] \[ \text{Area} = \frac{1}{2} |0| = 0 \] Alternatively, observe the slope between $(1,2)$ and $(3,6)$ is $\frac{6-2}{3-1} = 2$, and between $(3,6)$ and $(5,10)$ is $\frac{10-6}{5-3} = 2$. Since the slopes are equal, the points are collinear.

Step 4: Final Answer:
The area of the triangle is 0.

Answer 2

The area of a triangle formed by points \(A\), \(B\), and \(C\) can also be found using the magnitude of the cross product of two side vectors, \(\text{Area} = \dfrac{1}{2}|\vec{AB} \times \vec{AC}|\), instead of the coordinate determinant formula.

\[ \vec{AB} = (3-1,\ 6-2) = (2,4), \qquad \vec{AC} = (5-1,\ 10-2) = (4,8) \] \[ \vec{AB} \times \vec{AC} = (2)(8) - (4)(4) = 16 - 16 = 0 \]

1. 0: This matches the cross product calculation exactly, a zero cross product means the two side vectors point in exactly the same direction (one is a scalar multiple of the other), so no actual triangle is formed.
2. 2: A nonzero area like this would require the vectors \(\vec{AB}\) and \(\vec{AC}\) to point in genuinely different directions, but here \(\vec{AC} = 2\vec{AB}\) exactly, ruling out any nonzero spread between them.
3. 4: Similarly, this would require the cross product magnitude to be 8, but the calculation above gives exactly 0, leaving no room for this value.
4. 6: This would require a cross product magnitude of 12, again inconsistent with the direct computation showing the vectors are parallel.

Since \(\vec{AC}\) turns out to be precisely twice \(\vec{AB}\), the three points lie on a single straight line rather than forming a genuine triangle.

Therefore, the correct answer is 0.