A square has vertices at \(A(0,0)\), \(B(-9,12)\), \(C(3,21)\), and \(D(12,9)\).
The line \(\ell\) passes through \(A\) and intersects \(CD\) at point \(T(r,s)\), splitting the square so that the area of square \(ABCD\) is three times the area of \(\triangle ATD\).
Determine the equation of line \(\ell\).
Since \(A\) has coordinates \((0,0)\) and \(D\) has coordinates \((12, 9)\), using the distance formula, we have \[\begin{aligned} AD &= \sqrt{(9-0)^2 + (12-0)^2} \\ &= \sqrt{81 + 144}\\ & = \sqrt{225}\\ & = 15 \end{aligned}\] Therefore, the area of square \(ABCD\) is equal to \(15^2 = 225\).
Since the area of square \(ABCD\) is three times the area of \(\triangle ATD\), the area of \(\triangle ATD\) is equal to \(\frac{1}{3}\) of the area of square \(ABCD\). Thus, the area of \(\triangle ATD = \frac{1}{3}(225) = 75\).
Since \(ABCD\) is a square, \(\angle ADC = 90^{\circ}\). Consider \(\triangle ATD\). This triangle is a right-angled triangle with base \(AD = 15\) and height \(TD\).
Using the formula \(\text{area} = \dfrac{\text{base}\times\text{height}}{2}\), \[\begin{aligned} \text{area of }\triangle ATD &= \frac{AD\times TD}{2}\\ 75 &=\frac{15\times TD}{2}\\TD &=10 \end{aligned}\]
From here we present two solutions.
Solution 1
We first calculate the equation of the line that the segment \(CD\) lies on.
Since \(D\) has coordinates \((12,9)\) and \(C\) has coordinates \((3,21)\), this line has slope equal to \(\frac{21-9}{3-12} = \frac{12}{-9} = -\frac{4}{3}\).
Since the line has slope \(-\frac{4}{3}\) and the point \((3,21)\) lies on the line, we have \[\begin{aligned} \frac{y-21}{x-3} &= -\frac{4}{3} \\ 3y - 63 &= -4x +12 \\ 3y &= -4x + 75 \\ y &= -\frac{4}{3}x + 25 \end{aligned}\]
Since \(T(r,s)\) lies on this line, \(s = -\dfrac{4}{3}r + 25\).
Using the distance formula, since \(T\) has coordinates \((r,s)\), \(D\) has coordinates \((12,9)\), and \(TD = 10\), we have \[\begin{aligned} \sqrt{(r-12)^2 + (s-9)^2} &= 10\\ (r-12)^2 + (s-9)^2&= 100 \end{aligned}\]
Since \(s = -\frac{4}{3}r + 25\), we have \[\begin{aligned} (r-12)^2 + \left(\left(-\frac{4}{3}r + 25\right)-9\right)^2 &= 100\\ (r-12)^2 + \left(-\frac{4}{3}r+16\right)^2 &= 100\\ r^2 - 24r + 144 + \frac{16}{9}r^2 - \frac{128}{3}r + 256 &= 100\\ \frac{25}{9}r^2 - \frac{200}{3}r +300 &= 0\\ \frac{25}{9}(r^2 - 24r +108) &= 0\\ r^2 - 24r +108 &= 0\\ (r - 6)(r-18) &= 0\\ r & = 6, \ 18 \end{aligned}\] But \(r = 18\) lies outside the square. Therefore, \(r = 6\) and \(s = -\frac{4}{3}(6) + 25 = -8 + 25 = 17\).
Thus, the line \(\ell\) passes through \(A(0,0)\) and \(T(6, 17)\), has \(y\)-intercept \(0\), and slope \(\dfrac{17-0}{6-0} = \dfrac{17}{6}\).
Therefore, the equation of line \(\ell\) is \(y = \dfrac{17}{6}x\), or \(17x-6y=0\).
Solution 2
Since \(TD=10\) and \(CD=15\), we have \(CT=CD-TD=15-10=5\).
Since \(\triangle ATD\) is a right-angled triangle, using the Pythagorean Theorem we have \[\begin{aligned} AT^2&=AD^2+TD^2\\ (r-0)^2+(s-0)^2&=15^2+10^2\\ r^2+s^2&=325 \end{aligned}\]
Since \(T\) has coordinates \((r,s)\), \(C\) has coordinates \((3,21)\), and \(CT=5\), using the distance formula we have \[\begin{aligned} 5&=\sqrt{(r-3)^2+(s-21)^2}\\ 25&=r^2-6r+9+s^2-42s+441\\ 6r+42s&=r^2+s^2+425 \end{aligned}\] Since \(r^2+s^2=325\), we have \[\begin{aligned} 6r+42s&=325+425\\ &=750 \end{aligned}\]
Again, since \(T\) has coordinates \((r,s)\), \(D\) has coordinates \((12,9)\), and \(TD=10\), using the distance formula we have \[\begin{aligned} 10&=\sqrt{(r-12)^2+(s-9)^2}\\ 100&=r^2-24r+144+s^2-18s+81\\ 24r+18s&=r^2+s^2+125 \end{aligned}\] Since \(r^2+s^2=325\), we have \[\begin{aligned} 24r+18s&=325+125\\ &=450 \end{aligned}\]
We now have the system of equations \[\begin{aligned} 6r+42s &=750\\ 24r+18s&=450 \end{aligned}\] Multiplying the first equation by \(4\), we get the system \[\begin{aligned} 24r+168s &=3000\\ 24r+18s&=450 \end{aligned}\] Subtracting the second equation from the first gives \(150s=2550\), and \(s=17\) follows. Substituting \(s=17\) into \(6r+42s =750\), we obtain \(6r + 42(17) = 750\), and \(r=6\) follows.
Thus, the line \(\ell\) passes through \(A(0,0)\) and \(T(6, 17)\), has \(y\)-intercept \(0\), and slope \(\dfrac{17-0}{6-0} = \dfrac{17}{6}\).
Therefore, the equation of line \(\ell\) is \(y = \dfrac{17}{6}x\), or \(17x-6y=0\).
Extension:
Can you determine the coordinates of point \(U\) on \(CB\) such that the area of \(\triangle ABU\) is equal to the area of \(\triangle ATD\)? By finding \(U\) and \(T\), you will have found two line segments, \(AU\) and \(AT\), that divide square \(ABCD\) into three regions of equal area.