Question

select the correct answer.
what is the area of the triangle in the diagram?
a. $\frac{1}{2}sqrt{(x_{1}^{2}+y_{1}^{2})(x_{2}^{2}+y_{2}^{2})}$
b. $\frac{1}{2}sqrt{(x_{2}^{2}-x_{1}^{2})(y_{2}^{2}-y_{1}^{2})}$
c. $sqrt{(x_{1}^{2}+y_{1}^{2})(x_{2}^{2}+y_{2}^{2})}$
d. $2sqrt{(x_{2}^{2}-x_{1}^{2})(y_{2}^{2}-y_{1}^{2})}$

Explanation:

Step1: Recall the formula for the area of a right - angled triangle

The area of a right - angled triangle with two perpendicular sides \(a\) and \(b\) is \(A=\frac{1}{2}ab\). In a coordinate system, if we have two vectors \(\vec{v_1}=(x_1,y_1)\) and \(\vec{v_2}=(x_2,y_2)\) starting from the origin \((0,0)\), and the vectors are perpendicular, the lengths of the vectors are \(|\vec{v_1}|=\sqrt{x_1^{2}+y_1^{2}}\) and \(|\vec{v_2}|=\sqrt{x_2^{2}+y_2^{2}}\).

Step2: Calculate the area of the triangle

The area of the right - angled triangle formed by the two vectors starting from the origin is \(A = \frac{1}{2}|\vec{v_1}|\times|\vec{v_2}|\). Substituting the magnitudes of the vectors, we get \(A=\frac{1}{2}\sqrt{(x_1^{2}+y_1^{2})(x_2^{2}+y_2^{2})}\).

Answer:

A. \(\frac{1}{2}\sqrt{(x_1^{2}+y_1^{2})(x_2^{2}+y_2^{2})}\)