Question

△jkl is a right triangle with hypotenuse overline{kl}. leg overline{jl} is 2a units long, and leg overline{jk} is b units long. fill in the blanks to give the coordinates of the vertices of the triangle.

Explanation:

Step1: Analyze Vertex K

Assume the right angle is at \( J \) (since \( JK \) and \( JL \) are legs). If \( JK \) is horizontal (along y - axis? Wait, no, looking at the graph, \( JK \) seems horizontal (along the line from \( J \) to the y - axis? Wait, let's re - orient. Let's assume the right angle is at \( J \), so \( JK \) is one leg (length \( b \)) and \( JL \) is the other leg (length \( 2a \)). Let's set \( J \) at the origin? Wait, no, the graph has axes. Let's suppose:

- Let's assume \( J \) is at \( (0,0) \)? Wait, no, looking at the positions: \( K \) is along the y - axis? Wait, the diagram shows \( K \) above, \( J \) to the left, \( L \) at the bottom. Wait, maybe:

- Vertex \( K \): Let's say the vertical leg (along y - axis) has length related to \( b \)? Wait, maybe the coordinates:

- Let's assume the right angle is at \( J \). So \( JK \) is horizontal (length \( b \)) and \( JL \) is vertical (length \( 2a \)).

- So if \( J \) is at \( (0,0) \), then \( K \) would be at \( (0,b) \)? Wait, no, the horizontal line from \( J \) to the y - axis? Wait, maybe the correct approach is:

- For a right triangle \( \triangle JKL \) with hypotenuse \( KL \), legs \( JL = 2a \) and \( JK = b \). Let's assume the right angle is at \( J \). So:

- Let's set \( J \) at \( (0,0) \). Then, since \( JK \) is one leg (let's say horizontal), \( K \) would be at \( (0,b) \)? No, horizontal would be along x - axis. Wait, the diagram has a vertical axis (y) and horizontal axis (x). Let's re - interpret:

- The vertex \( K \): looking at the diagram, \( K \) is along the y - axis (the upper part). So let's say \( K \) has coordinates \( (0,b) \)? Wait, no, maybe:

- Let's assume the right angle is at \( J \), so \( JK \) is vertical (length \( b \)) and \( JL \) is horizontal (length \( 2a \)). Then:

- \( J \) is at \( (0,0) \), \( K \) is at \( (0,b) \), \( L \) is at \( (2a,0) \), and \( J \) is at \( (0,0) \). Wait, but the diagram shows \( J \) to the left, \( K \) above, \( L \) at the bottom.

- Wait, maybe the coordinates are:

- \( J \): Let's say \( J \) is at \( (0,0) \)

- \( K \): at \( (0,b) \) (since \( JK \) is vertical, length \( b \))

- \( L \): at \( (2a,0) \) (since \( JL \) is horizontal, length \( 2a \))

- And \( J \) is at \( (0,0) \)

- So filling the blanks:

- For \( K \): the x - coordinate is 0, y - coordinate is \( b \)? Wait, no, maybe the other way. Wait, the problem says "leg \( JL \) is \( 2a \) units long, and leg \( JK \) is \( b \) units long".

- Let's assume:

- \( J \) is at \( (0,0) \)

- \( K \) is at \( (0,b) \) (since \( JK \) is vertical, length \( b \))

- \( L \) is at \( (2a,0) \) (since \( JL \) is horizontal, length \( 2a \))

- And \( J \) is at \( (0,0) \)

- So the coordinates:

- \( J \): \( (0,0) \)

- \( K \): \( (0,b) \)

- \( L \): \( (2a,0) \)

- Wait, but the diagram shows \( J \) to the left (along the vertical axis? No, the left - most is \( J \), then the vertical axis, then \( K \) above.

- Alternatively, maybe \( J \) is at \( (0,0) \), \( K \) is at \( (0,b) \), \( L \) is at \( (2a,0) \), and \( J \) is at \( (0,0) \)

- So for \( K \): the x - coordinate is 0, y - coordinate is \( b \)

- For \( J \): x - coordinate is 0, y - coordinate is 0? No, the diagram shows \( J \) to the left of the vertical axis. Wait, maybe the correct coordinates are:

- \( J \): \( (0,0) \)

- \( K \): \( (0,b) \)

- \( L \): \( (2a,0) \)

- So filling the blanks:

- \( K \): \( (0,b) \)

- \( J \): \( (0,0) \) (but the diagram has two blanks for \( J \), maybe x and y? Wait, the blanks for \( J \): two boxes, so x and y coordinates. For \( K \): two boxes, x and y. For \( L \): two boxes.

- Let's re - do:

- Let's assume the right angle is at \( J \). So \( J \) has coordinates \( (0,0) \)

- \( JK \) is a leg of length \( b \), so if \( JK \) is vertical (along y - axis), then \( K \) is at \( (0,b) \)

- \( JL \) is a leg of length \( 2a \), so if \( JL \) is horizontal (along x - axis), then \( L \) is at \( (2a,0) \)

- So:

- \( J \): \( (0,0) \) (x = 0, y = 0)

- \( K \): \( (0,b) \) (x = 0, y = b)

- \( L \): \( (2a,0) \) (x = 2a, y = 0)

Step2: Verify the Triangle

- The distance between \( J(0,0) \) and \( K(0,b) \) is \( \sqrt{(0 - 0)^2+(b - 0)^2}=b \), which matches the length of leg \( JK \)

- The distance between \( J(0,0) \) and \( L(2a,0) \) is \( \sqrt{(2a - 0)^2+(0 - 0)^2}=2a \), which matches the length of leg \( JL \)

- The distance between \( K(0,b) \) and \( L(2a,0) \) is \( \sqrt{(2a - 0)^2+(0 - b)^2}=\sqrt{4a^{2}+b^{2}} \), which is the hypotenuse \( KL \), so this makes sense.

Answer:

• For \( J \): \( (0,0) \) (x - coordinate: \( 0 \), y - coordinate: \( 0 \))

• For \( K \): \( (0,b) \) (x - coordinate: \( 0 \), y - coordinate: \( b \))

• For \( L \): \( (2a,0) \) (x - coordinate: \( 2a \), y - coordinate: \( 0 \))

(Assuming the right angle is at \( J \) and the coordinate system as described above. The exact coordinates depend on the orientation of the triangle, but this is the most logical interpretation based on the given lengths and the diagram structure.)