Question

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. $2\sqrt{(x_{2}^{2}-x_{1}^{2})(y_{2}^{2}-y_{1}^{2})}$

Explanation:

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

The area of a right - angled triangle is \(A=\frac{1}{2}ab\), where \(a\) and \(b\) are the lengths of the two perpendicular sides.

Step2: Find the lengths of the two perpendicular sides

The length of one side (using the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), when one point is \((x_1,y_1)\) and the other is \((0,0)\)) is \(d_1=\sqrt{x_1^{2}+y_1^{2}}\). The length of the other side (when one point is \((x_2,y_2)\) and the other is \((0,0)\)) is \(d_2=\sqrt{x_2^{2}+y_2^{2}}\).

Step3: Calculate the area of the triangle

Substitute \(a = \sqrt{x_1^{2}+y_1^{2}}\) and \(b=\sqrt{x_2^{2}+y_2^{2}}\) into the area formula \(A=\frac{1}{2}ab\). We get \(A=\frac{1}{2}\sqrt{(x_1^{2}+y_1^{2})(x_2^{2}+y_2^{2})}\)

Answer:

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