In the adjoining figure, the area of triangle ABC is

1. 15 sq. units

2. 10 sq. units

3. 7.5 sq. units

4. 2.5 sq. units

Draw a perpendicular AD from A on x-axis.

From graph,

D = (1, 0).

By distance formula,

$d = \sqrt{(x$2 - x1)^2 + (y2 - y1)^2} \\[1em] \therefore AD = \sqrt{(1 - 1)^2 + (0 - 3)^2} \\[1em] = \sqrt{0 + (-3)^2} \\[1em] = \sqrt{9} \\[1em] = 3 \text{ units}. \\[1em] \therefore BC = \sqrt{[4 - (-1)]^2 + (0 - 0)^2} \\[1em] = \sqrt{[4 + 1]^2 + 0} \\[1em] = \sqrt{5^2} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.

Area of triangle = $\dfrac{1}{2} \times$ base × height

= $\dfrac{1}{2} \times 3 \times 5$

= $\dfrac{15}{2}$ = 7.5 sq. units

Hence, Option 3 is the correct option.