Right Triangle Hypotenuse: Endpoints At (1, 3)

Hey guys, ever stared at a geometry problem and thought, "What in the world am I supposed to do here?" Well, you're not alone! Today, we're diving deep into the fascinating world of right-angled triangles, specifically focusing on the hypotenuse and what happens when its endpoints are given as coordinates. We're talking about a right-angled triangle where the hypotenuse has its ends at the points (1, 3). This might sound a bit abstract at first, but stick with me, and by the end of this, you'll be a coordinate geometry whiz!

So, what exactly is a hypotenuse, you ask? In any right-angled triangle, the hypotenuse is that special, longest side that sits opposite the right angle. Think of it as the star of the show in any Pythagorean theorem exploration. When we're dealing with coordinates, these points are like little addresses on a graph. So, if the endpoints of our hypotenuse are at (1, 3), we're essentially looking at a line segment on a coordinate plane. This line segment is the hypotenuse of a right-angled triangle. The tricky part, or should I say the interesting part, is figuring out what kind of triangle this is, or what properties it has, given just these two points. It’s like having two crucial pieces of a puzzle and needing to deduce the rest of the picture. We’re not just finding the length of the hypotenuse here; we’re exploring the implications of those specific coordinates for the triangle itself. This means we’ll be pulling out our trusty distance formula, which is basically a souped-up version of the Pythagorean theorem for coordinates. Remember that formula? It's ${ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} }$. It’s your best friend when you need to find the distance between two points on a graph. For our points (1, 3), let's call them ${P_1 = (1, 3)}$ and ${P_2 = (x_2, y_2)}$. Wait a second, the prompt says the ends of the hypotenuse are at (1, 3). This implies that (1, 3) are two different points. Ah, a common hiccup! Let's correct that. The problem states the hypotenuse has its ends at the points (1, 3). This is ambiguous. It could mean:

1. The endpoints are P1(1, ?) and P2(?, 3).
2. The endpoints are P1(1, 3) and P2(some other point).
3. The problem intends to give two distinct points, and the notation (1, 3) is a typo for perhaps (1, y1) and (x2, 3) or even two distinct points like (1, a) and (b, 3) or (1, 3) and (4, 5).

Given the phrasing "has its ends at the points 1 3", it's most likely a typo and meant to imply two distinct points. Let's assume the prompt intended to give two distinct points, say ${A = (x_1, y_1)}$ and ${B = (x_2, y_2)}$, and that these are the endpoints of the hypotenuse. However, the prompt literally says "the points 1 3". This is highly unusual notation. It could mean the points are ${ (1, y_1) }$ and ${ (x_2, 3) }$, or perhaps it means the x-coordinates are 1 and 3, and the y-coordinates are also implied to be related or the same. This is where interpretation becomes key, and often, in math problems, clarity is king! Let's assume, for the sake of making progress and exploring the concept, that the prompt meant to provide two distinct points. If it literally means the points are ${P_1=(1, ?)}$ and ${P_2=(?, 3)}$, we can't proceed without more information. A more standard phrasing would be "the endpoints of the hypotenuse are at points A and B".

Let's consider the most likely interpretation of a slightly garbled prompt: that the two endpoints of the hypotenuse are distinct points, and perhaps the numbers '1' and '3' relate to their coordinates in some way. If we interpret "points 1 3" as referring to two specific points, say ${A = (1, y_A)}$ and ${B = (x_B, 3)}$, this still leaves us with unknowns. What if it meant the x-coordinates are 1 and 3? Like ${A = (1, y_A)}$ and ${B = (3, y_B)}$? Still too many unknowns.

The most probable interpretation, given typical geometry problems and the potential for typos, is that the prompt intended to give two specific points, say ${A = (x_1, y_1)}$ and ${B = (x_2, y_2)}$, and that these are the endpoints of the hypotenuse. The numbers "1 3" are likely a shorthand or error. If we must use "1 3" somehow, it's highly confusing. Could it mean the points are ${ (1, y_1) }$ and ${ (x_2, 3) }$ where ${ y_1 }$ and ${ x_2 }$ are also specified? Or maybe the points are ${ (1, 3) }$ and ${ (x_2, y_2) }$? This would mean one endpoint is at ${ (1, 3) }$, and the other is unknown. This doesn't fit "ends at the points 1 3" well.

Let's make a crucial assumption to move forward: The prompt is flawed and likely intended to give two coordinate pairs. Let's assume the endpoints of the hypotenuse are ${ A = (1, y_1) }$ and ${ B = (x_2, 3) }$. Even this is problematic as ${ y_1 }$ and ${ x_2 }$ are unknown.

A more plausible scenario for a solvable problem: The prompt meant to say something like, "the hypotenuse of a right-angled triangle has its endpoints at the points (1, 3) and (a, b)" where (a,b) are provided, OR, and this is a common type of problem, the vertices of the triangle are given, and one of them is the right angle, and the hypotenuse connects the other two. But here, it explicitly states the hypotenuse's ends are at (1, 3). This phrasing is the crux of our current dilemma!

Let's pivot to a common interpretation when such ambiguity arises: Perhaps "the points 1 3" is a very, very bad way of saying the endpoints are ${A = (1, y_A)}$ and ${B = (x_B, 3)}$ where some other information is provided, or maybe it means the x-coordinates are 1 and 3, and the y-coordinates are implied or identical? For example, if the points were ${A = (1, 5)}$ and ${B = (3, 5)}$, then the hypotenuse length is ${ extrm{distance} = extrm{sqrt}((3-1)^2 + (5-5)^2) = extrm{sqrt}(2^2) = 2 }$. This forms a horizontal line segment. If the points were ${A = (1, 5)}$ and ${B = (1, 3)}$, the hypotenuse length is ${ extrm{distance} = extrm{sqrt}((1-1)^2 + (3-5)^2) = extrm{sqrt}((-2)^2) = 2 }$. This forms a vertical line segment.

However, the prompt is specific about the endpoints of the hypotenuse being at "the points 1 3". This structure usually implies two distinct coordinate pairs. Given the context of geometry and coordinates, the most charitable interpretation to make this problem solvable and meaningful is that the prompt intended to provide two distinct points, and the notation "1 3" is either a typo for two coordinate pairs, or it's trying to convey something highly unconventional.

Let's imagine a scenario where the prompt actually meant: "The hypotenuse of a right-angled triangle has its endpoints at the points A(1, y1) and B(x2, 3)". Even here, without knowing ${y_1}$ and ${x_2}$, we can only calculate the length of the hypotenuse in terms of these variables: ${ L = extrm{sqrt}((x_2-1)^2 + (3-y_1)^2) }$. This isn't very satisfying.

The most common interpretation when faced with "points X Y" in a coordinate context is that X and Y represent coordinates themselves, not points. So, if the hypotenuse has ends at points (1, 3), it implies one endpoint is at (1, 3), and the other endpoint is missing. This contradicts "ends at the points 1 3" (plural).

Okay, let's take a deep breath and make the most reasonable assumption to proceed with a concrete example. We'll assume the prompt meant: "The hypotenuse of a right-angled triangle has its endpoints at the points A(1, y) and B(x, 3)", and then perhaps implies something about the third vertex. Or, even more likely, the prompt is missing information. Let's hypothesize the prompt intended to provide two specific points, and "1 3" is either a typo for two pairs of coordinates or refers to coordinates within those pairs.

Let's assume the prompt meant that the endpoints of the hypotenuse are the points ${ P_1 = (1, 3) }$ and ${ P_2 = (x_2, y_2) }$ and that the phrasing "the points 1 3" was a very confusing way to indicate one of the points is ${ (1, 3) }$ and perhaps the other point has coordinates derived from '1' and '3' in some way, or is just missing. This is pure speculation, of course! In a real exam, you'd ask for clarification. But for learning, we adapt!

Let's try a different angle: What if "the points 1 3" is a simplified reference to two points where the coordinates are implied? For instance, could it mean points ${A = (1, y_1)}$ and ${B = (x_2, 3)}$ where the other coordinates are maybe the same? Like ${A = (1, 3)}$ and ${B = (1, 3)}$? No, that's the same point. How about ${A = (1, k)}$ and ${B = (k, 3)}$? Still too open-ended.

The most constructive approach is to assume the prompt contained a typo and was intended to provide two distinct points. Let's say, for the sake of demonstration, that the endpoints of the hypotenuse are A(1, 5) and B(4, 3). Now we can actually do something!

First, calculate the length of the hypotenuse. This is fundamental. We use the distance formula we mentioned earlier: ${ d = extrm{sqrt}((x_2 - x_1)^2 + (y_2 - y_1)^2) }$.

Plugging in our assumed points A(1, 5) and B(4, 3): ${ extrm{Length of hypotenuse} = extrm{sqrt}((4 - 1)^2 + (3 - 5)^2) }$ ${ = extrm{sqrt}(3^2 + (-2)^2) }$ ${ = extrm{sqrt}(9 + 4) }$ ${ = extrm{sqrt}(13) }$

So, the length of our hypothetical hypotenuse is ${ extrm{sqrt}(13) }$ units. This is the direct result of having the endpoints defined. But why is this useful? Because the hypotenuse is a key component!

Now, remember, this is a right-angled triangle. This means there's a third vertex, let's call it C, such that the angle at C is 90 degrees, or the angle at A or B is 90 degrees. However, since AB is the hypotenuse, the right angle must be at the third vertex, C. So, we have vertices A(1, 5), B(4, 3), and an unknown vertex C(x, y), such that the lines AC and BC are perpendicular. This means the dot product of the vectors ${ extrm{AC} }$ and ${ extrm{BC} }$ is zero. Or, the product of their slopes is -1.

Let's explore the implications of the original prompt's wording one last time. If the hypotenuse has ends at "the points 1 3", and we are forced to interpret this literally as two points, it might mean: Point 1 is (1, ?) and Point 2 is (?, 3). Or maybe Point 1 is (1, 3) and Point 2 is (something else related to 1 and 3). This ambiguity is a classic example of how crucial clear problem statements are!

Let's consider the possibility that "1 3" refers to the coordinates of one point, and the prompt is incomplete. If one endpoint is ${ (1, 3) }$, and we need another endpoint, say ${ (x_2, y_2) }$, to define the hypotenuse. The length would be ${ extrm{sqrt}((x_2-1)^2 + (y_2-3)^2) }$. This is still not enough to define the triangle uniquely.

What if the prompt meant the set of coordinates involved are 1 and 3? This is a stretch, but let's think. Could the endpoints be ${ (1, 1) }$ and ${ (3, 3) }$? Or ${ (1, 3) }$ and ${ (3, 1) }$? If the endpoints were ${ A=(1, 3) }$ and ${ B=(3, 1) }$, then:

${ extrm{Length of hypotenuse} = extrm{sqrt}((3-1)^2 + (1-3)^2) }$ ${ = extrm{sqrt}(2^2 + (-2)^2) }$ ${ = extrm{sqrt}(4 + 4) }$ ${ = extrm{sqrt}(8) = 2 extrm{sqrt}(2) }$

This is a perfectly valid hypotenuse! Now, where would the right angle be? Let the third vertex be ${ C=(x, y) }$. The slopes of AC and BC must be negative reciprocals. Slope of AC = ${ (y-3)/(x-1) }$. Slope of BC = ${ (y-1)/(x-3) }$. For perpendicular lines, { rac{y-3}{x-1} imes rac{y-1}{x-3} = -1 }. This gives ${ (y-3)(y-1) = -(x-1)(x-3) }$, which simplifies to ${ y^2 - 4y + 3 = -(x^2 - 4x + 3) }$, or ${ x^2 - 4x + y^2 - 4y + 6 = 0 }$. This equation represents a circle centered at (2, 2) with radius ${ extrm{sqrt}(2) }$. Any point C on this circle forms a right angle at C with the hypotenuse AB. So, there are infinitely many possible right-angled triangles if the endpoints are (1, 3) and (3, 1).

The key takeaway here, guys, is that defining the endpoints of the hypotenuse is the first step. The distance formula is your go-to tool. Once you have the hypotenuse length, you can then explore the properties of the right-angled triangle, like finding the third vertex or calculating its area. However, the exact nature of the triangle is highly dependent on the exact coordinates of those endpoints. The ambiguity in "the points 1 3" is the main hurdle. If it were a clear two points, say ${ P_1(x_1, y_1) }$ and ${ P_2(x_2, y_2) }$, we could proceed with certainty. But as it stands, we've explored possibilities and highlighted the importance of precise mathematical language. Keep practicing, and don't be afraid to question ambiguous statements – that's a sign of a true mathematician in the making!

Calculating the Length of the Hypotenuse

Let's focus on the core task: finding the length of the hypotenuse given its endpoints. The distance formula is our trusty steed here. Remember it? It's derived directly from the Pythagorean theorem ${a^2 + b^2 = c^2}$. If we have two points, ${ (x_1, y_1) }$ and ${ (x_2, y_2) }$, we can imagine a right-angled triangle where the difference in x-coordinates ${ | extrm{x}_2 - extrm{x}_1| }$ is one leg, and the difference in y-coordinates ${ | extrm{y}_2 - extrm{y}_1| }$ is the other leg. The distance between the points is then the hypotenuse of this new right-angled triangle.

So, the distance ${d}$ is given by:

${ d = extrm{sqrt}(( extrm{x}_2 - extrm{x}_1)^2 + ( extrm{y}_2 - extrm{y}_1)^2) }$

Now, back to our prompt: "the hypotenuse of a right angled triangle has its ends at the points 1 3". As discussed, this notation is highly unusual and likely contains a typo. To make this section concrete, let's assume the prompt intended to provide two distinct points, and let's use the example we explored earlier where the endpoints are ${ A = (1, 3) }$ and ${ B = (3, 1) }$. These are two distinct points, and we can apply the distance formula.

Here, ${ x_1 = 1 }$, ${ y_1 = 3 }$ and ${ x_2 = 3 }$, ${ y_2 = 1 }$.

Plugging these values into the distance formula:

${ extrm{Length of Hypotenuse} = extrm{sqrt}((3 - 1)^2 + (1 - 3)^2) }$ ${ = extrm{sqrt}((2)^2 + (-2)^2) }$ ${ = extrm{sqrt}(4 + 4) }$ ${ = extrm{sqrt}(8) }$

We can simplify ${ extrm{sqrt}(8) }$ further. Since ${ 8 = 4 imes 2 }$, ${ extrm{sqrt}(8) = extrm{sqrt}(4 imes 2) = extrm{sqrt}(4) imes extrm{sqrt}(2) = 2 extrm{sqrt}(2) }$.

So, the length of the hypotenuse in this assumed scenario is ${ 2 extrm{sqrt}(2) }$ units. This calculation is straightforward once the endpoints are clearly defined. The challenge with the original prompt lies entirely in deciphering what "the points 1 3" actually refers to. If it meant one point was ${ (1, y_A) }$ and the other was ${ (x_B, 3) }$, the calculation would be ${ extrm{sqrt}((x_B - 1)^2 + (3 - y_A)^2) }$, which leaves us with unknown variables. This highlights how critical precise notation is in mathematics.

Finding the Third Vertex

Once we have the endpoints of the hypotenuse, say ${ A=(x_1, y_1) }$ and ${ B=(x_2, y_2) }$, the third vertex, ${ C=(x, y) }$, must form a right angle with A and B. This means the lines AC and BC are perpendicular. The condition for perpendicular lines (that are not horizontal or vertical) is that the product of their slopes is -1.

The slope of AC is { m_{AC} = rac{y - y_1}{x - x_1} }.

The slope of BC is { m_{BC} = rac{y - y_2}{x - x_2} }.

For perpendicularity:

${ m_{AC} imes m_{BC} = -1 }$

{ rac{y - y_1}{x - x_1} imes rac{y - y_2}{x - x_2} = -1 }

${ (y - y_1)(y - y_2) = -(x - x_1)(x - x_2) }$

${ y^2 - (y_1 + y_2)y + y_1y_2 = -(x^2 - (x_1 + x_2)x + x_1x_2) }$

${ y^2 - (y_1 + y_2)y + y_1y_2 = -x^2 + (x_1 + x_2)x - x_1x_2 }$

Rearranging gives:

${ x^2 - (x_1 + x_2)x + y^2 - (y_1 + y_2)y + x_1x_2 + y_1y_2 = 0 }$

This equation represents a circle. The center of this circle is at { ig( rac{x_1 + x_2}{2}, rac{y_1 + y_2}{2} ig) }, which is the midpoint of the hypotenuse. The radius squared is { ig(rac{x_1 - x_2}{2}ig)^2 + ig(rac{y_1 - y_2}{2}ig)^2 }, which is { rac{1}{4} ( extrm{length of hypotenuse})^2 }.

This means that the third vertex ${C}$ can be any point on the circle whose diameter is the hypotenuse AB. This tells us there are infinitely many right-angled triangles that can be formed with a given hypotenuse!

Let's use our assumed endpoints ${ A=(1, 3) }$ and ${ B=(3, 1) }$.

Here, ${ x_1 = 1, y_1 = 3, x_2 = 3, y_2 = 1 }$.

The equation becomes:

${ x^2 - (1 + 3)x + y^2 - (3 + 1)y + (1)(3) + (3)(1) = 0 }$

${ x^2 - 4x + y^2 - 4y + 3 + 3 = 0 }$

${ x^2 - 4x + y^2 - 4y + 6 = 0 }$

To find the center and radius, we complete the square:

${ (x^2 - 4x + 4) + (y^2 - 4y + 4) + 6 - 4 - 4 = 0 }$

${ (x - 2)^2 + (y - 2)^2 - 2 = 0 }$

${ (x - 2)^2 + (y - 2)^2 = 2 }$

This is a circle centered at ${ (2, 2) }$ with a radius of ${ extrm{sqrt}(2) }$. Any point ${ (x, y) }$ on this circle is a valid third vertex ${C}$ for our right-angled triangle.

For instance, if ${ x=2 }$, then ${ (2-2)^2 + (y-2)^2 = 2 }$, so ${ (y-2)^2 = 2 }$, which means ${ y-2 = extrm{pm} extrm{sqrt}(2) }$, giving ${ y = 2 extrm{pm} extrm{sqrt}(2) }$. So, ${ C = (2, 2+ extrm{sqrt}(2)) }$ or ${ C = (2, 2- extrm{sqrt}(2)) }$ are possible third vertices.

If ${ y=2 }$, then ${ (x-2)^2 + (2-2)^2 = 2 }$, so ${ (x-2)^2 = 2 }$, which means ${ x-2 = extrm{pm} extrm{sqrt}(2) }$, giving ${ x = 2 extrm{pm} extrm{sqrt}(2) }$. So, ${ C = (2+ extrm{sqrt}(2), 2) }$ or ${ C = (2- extrm{sqrt}(2), 2) }$ are other possible third vertices.

This demonstrates that knowing the hypotenuse endpoints defines the hypotenuse itself and a locus for the third vertex, but not a unique triangle unless additional constraints are provided. The ambiguity of the original prompt's "points 1 3" means we can only work with illustrative examples, but the principles remain the same!