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Problem of the Week
Problem E and Solution
The Hypotenuse is Aligned


\(\triangle OAB\) is an isosceles right-angled triangle with

Determine the area of \(\triangle OAB\).

Triangle A O B is plotted in the Cartesian plane. Vertex A lies in the second quadrant. Vertex B lies in the first quadrant.


Solution 1

Let \(B\) have coordinates \((p,q)\). Then the slope of \(OB = \frac{q}{p}\). Since \(\angle AOB=90^{\circ}\), then \(OB \perp OA\) and the slope of \(OA\) is the negative reciprocal of the slope of \(OB\). Therefore, the slope of \(OA = \frac{p}{-q}\). Since the triangle is isosceles with \(OA=OB\), it follows that the coordinates of \(A\) are \((-q,p)\). (We can verify this by finding the length of \(OA\) and the length of \(OB\) and showing that both lengths are equal to \(\sqrt {p^2+q^2}\).)

Since \(B(p,q)\) lies on the line \(2x+3y-13=0\), it satisfies the equation of the line.
Therefore, \(2p+3q-13=0\) (1).

Since \(A(-q,p)\) lies on the line \(2x+3y-13=0\), it satisfies the equation of the line.
Therefore, \(-2q+3p-13=0\), or \(3p-2q-13=0\) (2).

Since we have two equations and two unknowns, we can use elimination to solve for \(p\) and \(q\).

\[\begin{aligned} (1)\times 2:\quad 4p+6q-26&=0\\ (2)\times 3:\quad 9p-6q-39&=0\\ \text{Adding, we obtain}:\quad 13p-65&=0\\ p&=5\\ \text{Substituting in (1)}:\quad 10+3q-13&=0\\ 3q&=3\\ q&=1\end{aligned}\]

Therefore, the point \(B\) is \((5,1)\) and the length of \(OB\) is \(\sqrt{5^2+1^2}=\sqrt{26}\). Since \(OA=OB\), \(OA=\sqrt{26}\).
\(\triangle AOB\) is a right-angled triangle, so 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 \(\dfrac{OA \times OB}{2}=\dfrac{\sqrt{26}\sqrt{26}}{2}=13\).

Therefore, the area of \(\triangle AOB\) is \(13\text{ units}^2\).

Solution 2

By rearranging the given equation for the line, we obtain \(y=\frac{-2x+13}{3}\). Since the points \(A\) and \(B\) are 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 \(\triangle OAB\) is isosceles, we know that \(OA=OB\). Then

\[\begin{aligned} OA^2&=OB^2\\ a^2+\left(\frac{-2a+13}{3}\right)^2&=b^2+\left(\frac{-2b+13}{3}\right)^2\\ a^2+\frac{4a^2-52a+169}{9}&=b^2+\frac{4b^2-52b+169}{9}\\ \text{Multiplying by 9}:\quad 9a^2+4a^2-52a+169&=9b^2+4b^2-52b+169\\ \text{Simplifying}:\quad 13a^2-52a+169&=13b^2-52b+169\\ \text{Rearranging}:\quad 13a^2-13b^2-52a+52b&=0\\ \text{Dividing by 13}:\quad a^2-b^2-4a+4b&=0\\ \text{Factoring pairs}:\quad (a+b)(a-b)-4(a-b)&=0\\ \text{Common factoring}:\quad (a-b)(a+b-4)&=0 \end{aligned}\]

Solving, \(a=b\) or \(a=4-b\). Since \(A\) and \(B\) are distinct points, \(a\neq b\). Therefore, \(a=4-b\).
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 a right-angled triangle, we can use the Pythagorean Theorem, and \(AB^2=OA^2+OB^2\) follows. But \(OA=OB\), so this can be written \(AB^2=2OB^2\).

\[\begin{aligned} AB^2&=2OB^2\\ (b-(4-b))^2+\left(\frac{-2b+13}{3}-\frac{2b+5}{3}\right)^2&=2\left[b^2+\left(\frac{-2b+13}{3}\right)^2\right]\\ (2b-4)^2+\left(\frac{-4b+8}{3}\right)^2&=2\left[b^2+\frac{4b^2-52b+169}{9}\right]\\ 4b^2-16b+16+\frac{16b^2-64b+64}{9}&=2b^2+\frac{8b^2-104b+338}{9}\\ \text{Multiplying by 9}:\quad 36b^2-144b+144+16b^2-64b+64&=18b^2+8b^2-104b+338\\ \text{Simplifying}:\quad 52b^2-208b+208&=26b^2-104b+338\\ \text{Rearranging}:\quad 26b^2-104b-130&=0\\ \text{Dividing by 26}:\quad b^2-4b-5&=0\\ \text{Factoring}:\quad (b-5)(b+1)&=0\\\end{aligned}\]

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 \((-1,5)\). There are only two points. The area calculations shown in Solution 1 follow from here.

Therefore, the area of \(\triangle OAB\) is \(13\text{ units}^2\).