Peggie draws the line \(3x+4y=48\) and labels its intersection with the \(y\)-axis \(A\). Derrick then draws the line \(x+2y=22\) on the same set of axes and labels its intersection with Peggie’s line \(B\). Derrick then labels the intersection of his line with the \(x\)-axis \(C\). Peggie then labels the origin \(D\) and shades in quadrilateral \(ABCD\), as shown.
Determine the area of quadrilateral \(ABCD\).
We start by finding the coordinates of points \(A\), \(B\), and \(C\). Since \(D\) is the origin, its coordinates are \((0,0)\).
To determine the coordinates of point \(A\), we need to find the \(y\)-intercept of the line \(3x+4y=48\). To find this, let \(x=0\). Then \(4y=48\), so \(y=12\). Thus, the coordinates of \(A\) are \((0,12)\).
To determine the coordinates of point \(C\), we need to find the \(x\)-intercept of the line \(x+2y=22\). To find this, let \(y=0\). Then \(x=22\). Thus, the coordinates of \(C\) are \((22,0)\).
Point \(B\) is the point of intersection of the lines \(3x+4y=48\) and \(x+2y=22\). To find this, we solve the following system of equations. \[\begin{align} 3x+4y &= 48 \tag{1}\\ x + 2y &= 22 \tag{2} \end{align}\] From equation \((2)\), we solve for \(x\) to obtain \(x=22-2y\). Substituting into equation \((1)\), we have \[\begin{align} 3(22-2y) + 4y &=48\\ 66 - 6y + 4y&= 48\\ -2y &= -18\\ y&=9 \end{align}\] Then substituting back into \(x=22-2y\), we obtain \(x=22 - 2(9) = 4\). Thus, the coordinates of \(B\) are \((4,9)\).
We now draw line segment \(BD\) to divide quadrilateral \(ABCD\) into two triangles: \(\triangle ABD\) and \(\triangle CBD\) as shown.
\(\triangle ABD\) has a base of \(12\) and a height of \(4\), and \(\triangle CBD\) has a base of \(22\) and a height of \(9\). Thus, \[\begin{align} \text{Area }ABCD &= \text{Area }\triangle ABD + \text{Area }\triangle CBD\\ &= \frac{1}{2}(12)(4) + \frac{1}{2}(22)(9)\\ &= 24+99\\ &= 123 \end{align}\] Therefore, the area of quadrilateral \(ABCD\) is \(123 \text{ units}^2\).