A parabola intersects the \(y\)-axis at \(B(0,5)\), and intersects the \(x\)-axis at \(C(5,0)\) and at \(A(r,0)\), where \(0<r<5.\) The area of \(\triangle ABC\) is \(5\text{ units}^2.\)
If \(D(p,q)\) is the vertex of the parabola, then determine the area of \(\triangle DBC.\)
The height of \(\triangle ABC\) is the distance from the \(x\)-axis to \(B(0,5)\), which is \(5\) units. The base is \(AC=5-r.\) Since the area of \(\triangle ABC\) is \(5\), using the formula for the area of a triangle, we have \(\frac{(5-r)(5)}{2}=5.\) Then \(5-r=2\) and \(r=3\) follows. Thus, the coordinates of \(A\) are \((3,0).\)
The axis of symmetry of the parabola is a vertical line through the midpoint of \(AC\), which is \((4,0).\) It follows that the \(x\)-coordinate of the vertex is \(p=4.\) Therefore, the vertex is \(D(4,q).\)
Since the two \(x\)-intercepts of the parabola are \(3\) and \(5\), the equation of the parabola in factored form can be written as \(y=a(x-3)(x-5).\) Since the parabola passes through \(B(0,5)\), we can solve for \(a\) by substituting \(x=0\) and \(y=5\) into \(y=a(x-3)(x-5).\) This leads to \(a=\frac{1}{3}\) and thus the parabola has equation \(y=\frac{1}{3}(x-3)(x-5).\)
To determine \(q\), the \(y\)-coordinate of \(D\), we substitute \(x=4\), \(y=q\) into \(y=\frac{1}{3}(x-3)(x-5).\) Then \(q=\frac{1}{3}(4-3)(4-5)=-\frac{1}{3}.\) Therefore, \(D\) has coordinates \((4, -\frac{1}{3}).\)
From here, we proceed with two different solutions to determine the area of \(\triangle DBC.\)
Solution 1
Consider points \(E(0,-\frac{1}{3})\), \(F(5,-\frac{1}{3})\), and \(G(5,5)\), and draw in \(BGFE.\)
Since \(B\) and \(G\) have the same \(y\)-coordinate, \(BG\) is a horizontal line. Since \(G\) and \(F\) both have \(x\)-coordinate \(5\), \(GF\) is a vertical line which passes through \(C.\) Since \(E\) and \(F\) both have \(y\)-coordinate \(-\frac{1}{3}\), \(EF\) is a horizontal line which passes through \(D.\) Since \(B\) and \(E\) have the same \(x\)-coordinate, \(BE\) is a vertical line. Thus, \(BGFE\) is a rectangle that encloses \(\triangle DBC\), and we have \[\text{area }\triangle DBC = \text{area }BGFE - \text{area }\triangle BGC - \text{area }\triangle DFC - \text{area }\triangle BED\]
In rectangle \(BGFE\), \(BG=5-0=5\) and \(BE=5-(-\frac{1}{3})=\frac{16}{3}\). The area of rectangle \(BGFE=BG\times BE=5\times \frac{16}{3}=\frac{80}{3}\text{ units}^2.\)
Since \(BGFE\) is a rectangle, \(\triangle BGC\) is right-angled at \(G.\) Since \(BG=5\) and \(GC=5-0=5\), the area of \(\triangle BGC =\frac{BG\times GC}{2}=\frac{5\times 5}{2}=\frac{25}{2}\text{ units}^2.\)
Since \(BGFE\) is a rectangle, \(\triangle DFC\) is right-angled at \(F.\) Since \(CF=0-(-\frac{1}{3})=\frac{1}{3}\) and \(DF=5-4=1\), the area of \(\triangle DFC =\frac{CF\times DF}{2}=\frac{\frac{1}{3}\times 1}{2}=\frac{1}{6}\text{ units}^2.\)
Since \(BGFE\) is a rectangle, \(\triangle BED\) is right-angled at \(E.\) Since \(BE=\frac{16}{3}\) and \(ED=4-0=4\), the area of \(\triangle BED =\frac{BE\times ED}{2}=\frac{\frac{16}{3}\times 4}{2}=\frac{32}{3}\text{ units}^2.\)
Thus, \[\begin{aligned} \text{area }\triangle DBC &= \text{area }BGFE - \text{area }\triangle BGC - \text{area }\triangle DFC - \text{area }\triangle BED\\ &=\tfrac{80}{3} - \tfrac{25}{2} - \tfrac{1}{6} - \tfrac{32}{3}\\ &=\tfrac{10}{3}\text{ units}^2 \end{aligned}\]
Solution 2
Let \(P(t,0)\) be the point where the line through \(B\) and \(D\) crosses the \(x\)-axis. We will determine the equation of the line that passes through \(B\), \(P\), and \(D.\)
Since the line passes through \(B(0,5)\) and \(D(4,-\frac{1}{3})\), the slope of the line is \(\frac{5+\frac{1}{3}}{0-4}=\frac{\frac{16}{3}}{-4}=-\frac{4}{3}.\)
The \(y\)-intercept of the line is \(5.\) Therefore, the equation of the line through \(B\), \(P\), and \(D\) is \(y=-\frac{4}{3}x+5.\)
To determine \(t\), the \(x\)-coordinate of \(P\) we substitute \(x=t\) and \(y=0\) into \(y=-\frac{4}{3}x+5\), the equation of the line. Thus, \(0=-\frac{4}{3}t+5\), and \(4t=15\) or \(t=\frac{15}{4}\) follows.
In \(\triangle BPC\), the height is the perpendicular distance from the \(x\)-axis to point \(B\), which is \(5.\) The base is \(PC = 5-\frac{15}{4}=\frac{5}{4}.\) Thus, the area of \(\triangle BPC =\frac{\frac{5}{4}\times 5}{2}=\frac{25}{8}\text{ units}^2.\)
In \(\triangle DPC\), the height is the perpendicular distance from the \(x\)-axis to point \(D\), which is \(\frac{1}{3}\). The base is \(PC = 5-\frac{15}{4}=\frac{5}{4}.\) Thus, the area of \(\triangle DPC =\frac{\frac{5}{4}\times \frac{1}{3}}{2}=\frac{5}{24}\text{ units}^2.\)
Therefore, the area of \(\triangle DBC=\text{area }\triangle BPC+\text{area }\triangle DPC = \frac{25}{8}+\frac{5}{24} = \tfrac{10}{3}\text{ units}^2.\)