Points \(A(0,a)\), \(B(2,-1)\), \(C(3,2)\), \(D(0,-1)\), and \(O(0,0)\) are such that \(\triangle ABD\) and \(\triangle COB\) have the same area. If \(a>0\), determine the value of \(a.\)
Solution 1
In \(\triangle ABD\), \(AD = a - (-1) = a+1\) and \(DB = 2-0 = 2.\)
Thus, area \(\triangle ABD = \frac{AD\times DB}{2} = \frac{(a+1)\times 2}{2} = a+1 \text{ units}^2.\)
To determine the area of \(\triangle COB\), we consider points \(E(0,2)\) and \(F(3,-1)\) and draw in \(ECFD\).
Since \(E\) and \(D\) both have \(x\)-coordinate \(0\), \(ED\) is a vertical line which passes through \(O.\) Since \(C\) and \(F\) have the same \(x\)-coordinate, \(CF\) is also a vertical line. Since \(E\) and \(C\) have the same \(y\)-coordinate, \(EC\) is a horizontal line. Since \(D\) and \(F\) both have \(y\)-coordinate \(-1\), \(DF\) is also a horizontal line which passes through \(B.\) Thus, \(ECFD\) is a rectangle that encloses \(\triangle COB\), and we have \[\text{area }\triangle COB = \text{area }ECFD - \text{area }\triangle CEO - \text{area }\triangle ODB - \text{area }\triangle BFC\]
In rectangle \(ECFD\), \(EC=3-0=3\) and \(ED=2-(-1)=3\). The area of rectangle \(EDFC=EC\times ED=3\times 3=9\text{ units}^2.\)
Since \(ECFD\) is a rectangle, \(\triangle CEO\) is right-angled at \(E.\) Since \(EC=3\) and \(EO=2-0=2\), the area of \(\triangle CEO =\frac{EC\times EO}{2}=\frac{3\times 2}{2}=3\text{ units}^2.\)
Since \(ECFD\) is a rectangle, \(\triangle ODB\) is right-angled at \(D.\) Since \(OD=0-(-1)=1\) and \(DB=2-0=2\), the area of \(\triangle ODB =\frac{OD\times DB}{2}=\frac{1\times 2}{2}=1\text{ unit}^2.\)
Since \(ECFD\) is a rectangle, \(\triangle BFC\) is right-angled at \(F.\) Since \(BF=3-2=1\) and \(CF=2-(-1)=3\), the area of \(\triangle BFC =\frac{BF\times CF}{2}=\frac{1\times 3}{2}=1.5\text{ units}^2.\)
Thus, \[\begin{aligned} \text{area }\triangle COB&=\text{area }ECFD - \text{area }\triangle CEO - \text{area }\triangle ODB - \text{area }\triangle BFC\\ &=9-3-1-1.5\\ &=3.5\text{ units}^2 \end{aligned}\]
We’re given that \(\triangle ABD\) and \(\triangle COB\) have the same area. Thus, the area of \(\triangle ABD=3.5\text{ units}^2.\)
Since the area of \(\triangle ABD = a+1\text{ units}^2\), we have \(a+1 = 3.5\) and \(a=2.5\) follows.
Therefore, the value of \(a\) is \(2.5.\)
Solution 2
In \(\triangle ABD\), \(AD = a - (-1) = a+1\) and \(DB = 2-0 = 2.\)
Thus, area \(\triangle ABD = \frac{AD\times DB}{2} = \frac{(a+1)\times 2}{2} = a+1 \text{ units}^2.\)
Let \(P(p,0)\) be the point where the line through \(C(3,2)\) and \(B(2,-1)\) intersects the \(x\)-axis.
We have \(\text{area }\triangle COB = \text{area }\triangle COP + \text{area }\triangle BOP.\)
To determine the value of \(p\), we first determine the equation of the line through \(C(3,2)\) and \(B(2,-1).\)
Since the slope of the line is \(\frac{2-(-1)}{3-2}=3\), the equation of the line is of the form \(y=3x+b\), for some \(b.\) Substituting \(x=3\) and \(y=2\) gives \(2=3(3)+b\) and \(b=-7\) follows. Therefore, the equation of the line though \(C(3,2)\) and \(B(2,-1)\) is \(y=3x-7.\)
Substituting \(x=p\) and \(y=0\) into \(y=3x-7\) we obtain \(0=3p-7\) and \(p=\frac{7}{3}\) follows.
In \(\triangle COP\), \(OP=\frac{7}{3}\) and the height is the perpendicular distance from the \(x\)-axis to \(C(3,2)\), which is 2 units. Therefore, the area of \(\triangle COP=\frac{\frac{7}{3}\times 2}{2}=\frac{7}{3}\text{ units}^2.\)
In \(\triangle BOP\), \(OP=\frac{7}{3}\) and the height is the perpendicular distance from the \(x\)-axis to \(B(2,-1)\), which is 1 unit. Therefore, the area of \(\triangle BOP=\frac{\frac{7}{3}\times 1}{2}=\frac{7}{6}\text{ units}^2.\)
Therefore, \(\text{area }\triangle COB = \text{area }\triangle COP + \text{area }\triangle BOP = \frac{7}{3} + \frac{7}{6} = \frac{14}{6}+\frac{7}{6} = \frac{21}{6} = \frac{7}{2}\text{ units}^2.\)
We’re given that \(\triangle ABD\) and \(\triangle COB\) have the same area. Thus, the area of \(\triangle ABD=\frac{7}{2}\text{ units}^2.\)
Since the area of \(\triangle ABD = a+1\text{ units}^2\), we have \(a+1 = \frac{7}{2}\) and \(a=\frac{5}{2} = 2.5\) follows.
Therefore, the value of \(a\) is \(2.5.\)