Area of the triangle having the coordinates (4,2),(-1,2) and (3,b) is 10 sq.uints. Find the value of b?

Concept: The area of a triangle is $\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$ where $\left(x_1,y_1\right),\left(x_2,y_2\right)$ and $\left(x_3,y_3\right)$ are the vertices of the triangle.

Here, the vertices of the triangle are  $(4,2),(-1,2)$ and $(3, b)$.

We know, $area of \Delta ABC=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$.

Area of the triangle is 10 square units.

$$\begin{gathered} \Rightarrow \frac{1}{2}\left[4\left(2-b\right)+(-1)\left(b-2\right)+3\left(2-2\right)\right]=10 \\ \Rightarrow \frac{1}{2}\left[8-4b-b+2\right]=10 \\ \Rightarrow \frac{1}{2}[-5b+10]=10 \\ \Rightarrow \frac{-5b}{2}=5 \\ \Rightarrow b=-2 \end{gathered}$$

Hence, the value of $b$ is $-2.$