The area of a triangle is 5 sq. units. Two of its vertices are at $(2, 1)$ and $(3, -2)$. If the third vertex is $(\frac{7}{2} , y)$, find $y$.

Given:

The vertices of a triangle are $(2, 1), (3, -2)$ and $(\frac{7}{2}, y)$ and its area is 5 sq. units.

To do:

We have to find the value of $y$.

Solution:

Let $A(2, 1), B(3, -2)$ and $C(\frac{7}{2}, y)$ be the vertices of $\triangle ABC$.

We know that,

Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by, 

Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

Therefore,

Area of triangle \( ABC=\frac{1}{2}[2(-2-y)+3(y-1)+\frac{7}{2}(1+2)] \)

\( \pm 5=\frac{1}{2}[-4-2y+3y-3+\frac{7}{2}(3)] \)

\( \pm 5(2)=(y-7+\frac{21}{2}) \)

\( \pm 10+7-\frac{21}{2}=y \)

\( y=\frac{2(17)-21}{2} \) or \( y=\frac{2(-3)-21}{2} \)

\( y=\frac{34-21}{2} \) or \( y=\frac{-6-21}{2} \)

\( y=\frac{13}{2} \) or \( y=\frac{-27}{2} \)

The value of $y$ is $\frac{13}{2}$ or $\frac{-27}{2}$.