Chapter 8.5, Problem 20E

To determine

The value of $y$ such that the triangle with given vertices $\left(-4,2\right),\left(-3,5\right),\left(-1,y\right)$ has an area of $4$ square units.

Answer to Problem 20E

The value $\text{'}y\text{'}$ of is $0or\frac{16}{5}$ such that the area of triangle is $4$ square units.

Explanation of Solution

Given information:

Given vertices $\left(-4,2\right),\left(-3,5\right),\left(-1,y\right)$ of triangle.

Area of triangle $A$ with veritcies $\left(x_1,y_1\right),\left(x_2,y_2\right),\left(x_3,y_3\right)$ is given by $\left|A\right|=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

As given area is $4$ square units.

Thus, $\left|4\right|=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

  $\begin{array}{l}\Rightarrow \pm 8=\left|\begin{array}{ccc}-4& 2& 1\\ -3& 5& 1\\ -1& y& 1\end{array}\right|\\ \Rightarrow \pm 8=\left(-4\right)\left(5-y\right)-2\left(-3+1\right)+1\left(-3y+5\right)\\ \Rightarrow \pm 8=-20+4y+4-3y+5\\ \Rightarrow \pm 8=-11+y\\ \Rightarrow -8=-11+yor8=-11+5y\\ \Rightarrow y=3ory=19\\ \Rightarrow y=3ory=19.\end{array}$

Therefore, the value $\text{'}y\text{'}$ of is $3or19$ such that the area of triangle is $4$ square units.