Suppose that \(\triangle OAB\) is such that vertex \(O\) is located at the origin, and vertices \(A\) and \(B\) lie on the line \(2x+3y-13=0\) with \(\angle AOB=90\degree\) and \(OA=OB\).
Determine the area of \(\triangle OAB\).
By rearranging the given equation for the line, we obtain \(y=\frac{-2x+13}{3}\).
Since points \(A\) and \(B\) lie on the line, their coordinates satisfy the equation of the line. If \(A\) has \(x\)-coordinate \(a\), then \(A\) has coordinates \(\left(a,\frac{-2a+13}{3}\right)\). If \(B\) has \(x\)-coordinate \(b\), then \(B\) has coordinates \(\left(b,\frac{-2b+13}{3}\right)\).
Since \(OA=OB\), we have \[\begin{aligned} OA^2&=OB^2\\ a^2+\left(\frac{-2a+13}{3}\right)^2&=b^2+\left(\frac{-2b+13}{3}\right)^2\\ a^2+\frac{4a^2-52a+169}{9}&=b^2+\frac{4b^2-52b+169}{9}\\ 9a^2+4a^2-52a+169&=9b^2+4b^2-52b+169\\ 13a^2-52a+169&=13b^2-52b+169\\ 13a^2-13b^2-52a+52b&=0\\ a^2-b^2-4a+4b&=0 \end{aligned}\] By factoring, we then obtain \((a+b)(a-b)-4(a-b)=0\) or \((a-b)(a+b-4)=0\).
This implies that, \(a=b\) or \(a=4-b\). Since \(A\) and \(B\) are distinct points, \(a\neq b\). Therefore, \(a=4-b\).
From here we proceed with two different solutions.
Solution 1
Since \(B\) has coordinates \(\left(b,\frac{-2b+13}{3}\right)\), the slope of \(OB\) is equal to \(\dfrac{\frac{-2b+13}{3}}{b}= \dfrac{-2b + 13}{3b}\).
Since \(A\) has coordinates \(\left(a,\frac{-2a+13}{3}\right)\), the slope of \(OA\) is equal to \(\dfrac{\frac{-2a+13}{3}}{a}= \dfrac{-2a + 13}{3a}\).
Since \(\angle AOB=90^\circ\), then \(OB \perp OA\) and the slope of \(OA\) is the negative reciprocal of the slope of \(OB\). Therefore, \[\begin{aligned} \dfrac{-2a + 13}{3a} &= -\dfrac{1}{\dfrac{-2b + 13}{3b}}\\ &= \dfrac{-3b}{-2b+13} \end{aligned}\] Simplifying, we obtain \[\begin{aligned} (-2a+13)(-2b+13) &= (-3b)(3a)\\ 4ab - 26a - 26b +169 &=-9ab\\ 13ab -26a - 26b + 169&=0\\ ab -2a-2b+13 &=0 \end{aligned}\]
Substituting \(a=4-b\), we have \[\begin{aligned} (4-b)b-2(4-b) -2b +13 &=0\\ 4b-b^2-8+2b-2b+13 &=0\\ b^2 -4b -5&=0 \end{aligned}\] Factoring, this becomes \((b-5)(b+1)=0\). It follows that \(b=5\) or \(b=-1\). When \(b=5\), the point \(A\) is \((-1,5)\) and the point \(B\) is \((5,1)\).
When \(b=-1\), the point \(A\) is \((5,1)\) and the point \(B\) is \((-1,5)\).
In each case, the length of \(OB\) is \(\sqrt{5^2+1^2}=\sqrt{26}\). Since \(OA=OB\), \(OA=\sqrt{26}\).
Since \(\triangle AOB\) is a right-angled triangle, we can use \(OB\) as the base and \(OA\) as the height in the formula for the area of a triangle.
Therefore, the area of \(\triangle AOB\) is \(\dfrac{OA \times OB}{2}=\dfrac{\sqrt{26}\times\sqrt{26}}{2}=13\).
Therefore, the area of \(\triangle AOB\) is \(13\text{ units}^2\).
Solution 2
Since \(a=4-b\), we can rewrite \(A\left(a,\frac{-2a+13}{3}\right)\) as \(A\left(4-b,\frac{-2(4-b)+13}{3}\right)\) which simplifies to \(A\left(4-b,\frac{2b+5}{3}\right)\).
Since \(\triangle OAB\) is right-angled at \(O\), by the Pythagorean Theorem, \(AB^2=OA^2+OB^2\). Since \(OA=OB\), this can be written \(AB^2=2OB^2\). Thus, \[\begin{aligned} (b-(4-b))^2+\left(\frac{-2b+13}{3}-\frac{2b+5}{3}\right)^2&=2\left(b^2+\left(\frac{-2b+13}{3}\right)^2\right)\\ (2b-4)^2+\left(\frac{-4b+8}{3}\right)^2&=2\left(b^2+\frac{4b^2-52b+169}{9}\right)\\ 4b^2-16b+16+\frac{16b^2-64b+64}{9}&=2b^2+\frac{8b^2-104b+338}{9} \end{aligned}\] Multiplying by \(9\) and simplifying, we obtain \[\begin{aligned} 36b^2-144b+144+16b^2-64b+64&=18b^2+8b^2-104b+338\\ 52b^2-208b+208&=26b^2-104b+338\\ 26b^2-104b-130&=0\\ b^2-4b-5&=0 \end{aligned}\] Factoring, this becomes \((b-5)(b+1)=0\). It follows that \(b=5\) or \(b=-1\). When \(b=5\), the point \(A\) is \((-1,5)\) and the point \(B\) is \((5,1)\).
When \(b=-1\), the point \(A\) is \((5,1)\) and the point \(B\) is \((-1,5)\).
In each case, the length of \(OB\) is \(\sqrt{5^2+1^2}=\sqrt{26}\). Since \(OA=OB\), \(OA=\sqrt{26}\).
Since \(\triangle AOB\) is a right-angled triangle, we can use \(OB\) as the base and \(OA\) as the height in the formula for the area of a triangle.
Therefore, the area of \(\triangle AOB\) is \(\dfrac{OA \times OB}{2}=\dfrac{\sqrt{26}\times\sqrt{26}}{2}=13\).
Therefore, the area of \(\triangle AOB\) is \(13\text{ units}^2\).