build perseverance find the coordinates of point l so △jkl is a right triangle. point j has coordinates (0, 0), and point k has coordinates (2a, 2b).

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# Explanation:
## Step1: Analyze right triangle cases
A right triangle can have the right angle at $ J $, $ K $, or $ L $.
- **Case 1: Right angle at $ J(0,0) $**
  If $ \angle J = 90^\circ $, then $ JL $ is vertical/horizontal and $ JK $ is the other leg. Since $ J(0,0) $ and $ K(2a,2b) $, a perpendicular line at $ J $ could be along the x - axis or y - axis. A simple choice is $ L(2a,0) $ (here, $ JL $ is horizontal ($ y = 0 $) and $ JK $ has slope $ \frac{2b - 0}{2a - 0}=\frac{b}{a} $, and $ JL $ has slope $ 0 $, $ JK $ and $ JL $ are perpendicular when $ a
eq0 $; or $ L(0,2b) $ (similar logic with vertical line)). Let's take $ L(2a,0) $ for simplicity. We can verify the slopes: slope of $ JK=\frac{2b - 0}{2a - 0}=\frac{b}{a} $, slope of $ JL=\frac{0 - 0}{2a - 0}=0 $, slope of $ KL=\frac{0 - 2b}{2a - 2a} $ (undefined, but actually, when $ L(2a,0) $, $ JL $ is horizontal, $ JK $ is a line from $ (0,0) $ to $ (2a,2b) $, and $ KL $ is vertical. The product of slopes of $ JL $ (slope $ 0 $) and $ JK $ (slope $ \frac{b}{a} $) is $ 0\times\frac{b}{a} = 0$, but actually, for right angle at $ J $, the vectors $ \overrightarrow{JJ}=(0,0) $ (no, better to use vectors $ \overrightarrow{JL}=(x, y) $ and $ \overrightarrow{JK}=(2a,2b) $. Their dot product should be zero: $ \overrightarrow{JL}\cdot\overrightarrow{JK}=2a\times x+2b\times y = 0 $. If we take $ x = 2a $, $ y = 0 $, then $ 2a\times2a+2b\times0 = 4a^{2}=0 $ only if $ a = 0 $, which is not general. Wait, maybe a better approach:
- **Case 2: Right angle at $ K(2a,2b) $**
  Vectors $ \overrightarrow{KJ}=(- 2a,-2b) $ and $ \overrightarrow{KL}=(x - 2a,y - 2b) $. Their dot product should be zero: $ - 2a(x - 2a)-2b(y - 2b)=0\Rightarrow - 2ax + 4a^{2}-2by + 4b^{2}=0\Rightarrow ax+by=2a^{2}+2b^{2} $. A simple solution is $ L(0,2b) $: check dot product $ \overrightarrow{KJ}=(-2a,-2b) $, $ \overrightarrow{KL}=(-2a,0) $, dot product $ (-2a)\times(-2a)+(-2b)\times0 = 4a^{2}
eq0 $ (wrong). Wait, maybe the easiest cases are when $ L $ is such that either $ JL $ is horizontal and $ KL $ is vertical or vice - versa.
- **Case 3: Right angle at $ L $**
  Let $ L=(x,y) $. Then $ \overrightarrow{JL}=(x,y) $, $ \overrightarrow{KL}=(x - 2a,y - 2b) $, and $ \overrightarrow{JK}=(2a,2b) $. For right angle at $ L $, $ \overrightarrow{JL}\cdot\overrightarrow{KL}=0\Rightarrow x(x - 2a)+y(y - 2b)=0 $. But the simplest non - trivial cases are when $ L $ is $ (2a,0) $ or $ (0,2b) $ (right angle at $ K $ or $ J $ respectively). Let's take $ L=(2a,0) $:
  - $ J(0,0) $, $ K(2a,2b) $, $ L(2a,0) $
  - Length of $ JL=\sqrt{(2a - 0)^{2}+(0 - 0)^{2}} = 2|a| $
  - Length of $ KL=\sqrt{(2a - 2a)^{2}+(0 - 2b)^{2}} = 2|b| $
  - Length of $ JK=\sqrt{(2a - 0)^{2}+(2b - 0)^{2}} = 2\sqrt{a^{2}+b^{2}} $
  - Check Pythagoras: $ JL^{2}+KL^{2}=(2a)^{2}+(2b)^{2}=4a^{2}+4b^{2} $, and $ JK^{2}=(2\sqrt{a^{2}+b^{2}})^{2}=4(a^{2}+b^{2}) $. So $ JL^{2}+KL^{2}=JK^{2} $, so $ \triangle JKL $ is right - angled at $ L $.

Another simple solution is $ L=(0,2b) $. Let's verify:
- $ J(0,0) $, $ K(2a,2b) $, $ L(0,2b) $
- $ JL=\sqrt{(0 - 0)^{2}+(2b - 0)^{2}} = 2|b| $
- $ KL=\sqrt{(0 - 2a)^{2}+(2b - 2b)^{2}} = 2|a| $
- $ JK = 2\sqrt{a^{2}+b^{2}} $
- $ JL^{2}+KL^{2}=4b^{2}+4a^{2}=JK^{2} $, so it is also a right triangle.

## Step2: Conclusion
One of the possible coordinates of $ L $ is $ (2a,0) $ (or $ (0,2b) $). We will take $ (2a,0) $ as a valid solution (other solutions like $ (0,2b) $ also exist).

# Answer:
$(2a,0)$ (or $(0,2b)$ is also a valid answer)

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Step1: Analyze right triangle cases

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