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

Problem

$$\triangle OAB$$ is an isosceles right-angled triangle with

• vertex $$O$$ located at the origin; and

• vertices $$A$$ and $$B$$ located on the line $$2x+3y-13=0$$ such that $$\angle AOB=90^\circ$$ and $$OA=OB$$.

Determine the area of $$\triangle OAB$$.

Solution

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 $$13\text{ units}^2$$.