College Algebra, Chapter 2, 2.4, Section 2.4, Problem 62

Determine the area of the triangle formed by the coordinates axes and the line $2y + 3x - 6 = 0$

Solving for $y$ intercept, where $x = 0$


$
\begin{equation}
\begin{aligned}

2y + 3(0) - 6 =& 0
\\
\\
2y - 6 =& 0
\\
\\
2y =& 6
\\
\\
y =& 3

\end{aligned}
\end{equation}
$


The $y$ intercept is at $(0,3)$

Solving for $x$ intercept, where $y = 0$


$
\begin{equation}
\begin{aligned}

2(0) + 3x - 6 =& 0
\\
\\
3x - 6 =& 0
\\
\\
3x =& 6
\\
\\
x =& 2

\end{aligned}
\end{equation}
$


The $x$ intercept is at $(2, 0)$

Since the $y$ intercept is at $(0,3)$, notice that the height of the triangle will be $h = 3$. Consequently, if the $x$ intercept is at $(2, 0)$, then the base of the triangle will be $b = 2$. Thus, the area is..

$\displaystyle A = \frac{1}{2} bh = \frac{1}{2} (2)(3) = 3 $ square units