What is the area of the triangle whose vertices are (0,1), (0,2), (1,5)?

The correct option is 1/2 sq. units.

To find the area of a triangle with given vertices, we can use the coordinate geometry formula for the area of a triangle. Let the vertices of the triangle be:

$$(x_1,y_1)=(0,1)$$

$$(x_2,y_2)=(0,2)$$

$$(x_3,y_3)=(1,5)$$

The formula for the area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ is given by:

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

Now, let's substitute the coordinates of our vertices into this formula:
Substitute $x_1=0$, $y_1=1$, $x_2=0$, $y_2=2$, $x_3=1$, and $y_3=5$:

$$\text{Area}=\frac{1}{2}|0\cdot (2-5)+0\cdot (5-1)+1\cdot (1-2)|$$

Simplifying the terms inside the absolute value brackets:
The first and second terms simplify to 0 because they are multiplied by 0:

$$\text{Area}=\frac{1}{2}|0+0+1\cdot (-1)|$$

This further simplifies to:

$$\text{Area}=\frac{1}{2}|-1|$$

Since the absolute value of $-1$ is $1$, we have:

$$\text{Area}=\frac{1}{2}\cdot 1=\frac{1}{2}\text{ sq. units}$$

Thus, the area of the triangle is indeed 1/2 sq. units.