- #1

Wilmer

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BC = a = sqrt(17)

AC = b = sqrt(68)

AB = c = sqrt(85)

A's coordinates: 0,12

B's coordinates: 6,5

What's EASIEST way to get C's coordinates?

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- Thread starter Wilmer
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In summary, a right triangle ABC is given with side lengths BC = a = sqrt(17), AC = b = sqrt(68), AB = c = sqrt(85). The coordinates of points A and B are given as (0,12) and (6,5) respectively. The easiest way to find the coordinates of point C is by realizing that C will lie on a circle with AB as its diameter. There will be two possible answers for C, on either side of AB. The two solutions for C are (2,4) and (38/5,44/5). However, if trigonometry is not used, the problem can be solved by observing that triangles ABC and ACD are similar, meaning that they

- #1

Wilmer

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BC = a = sqrt(17)

AC = b = sqrt(68)

AB = c = sqrt(85)

A's coordinates: 0,12

B's coordinates: 6,5

What's EASIEST way to get C's coordinates?

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- #2

Olinguito

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- #3

Wilmer

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And the 2 solutions for C are (2,4) and (38/5,44/5).

BUT I was trying to solve without involving a circle.

(guess I should have said so!)

Can you tell me it's definitely impossible?

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I like Serena

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Wilmer said:

BC = a = sqrt(17)

AC = b = sqrt(68)

AB = c = sqrt(85)

A's coordinates: 0,12

B's coordinates: 6,5

What's EASIEST way to get C's coordinates?

Let's make a drawing:

\begin{tikzpicture}[>=stealth']

%preamble \usetikzlibrary{arrows}

\def\a{sqrt(17)}

\def\b{sqrt(68)}

\def\c{sqrt(85)}

\coordinate (A) at (0,12);

\coordinate (B) at (6,5);

\coordinate (C1) at (2,4);

\coordinate (C2) at ({38/5},{44/5});

\coordinate (D) at ({\b*\b/\c/\c * 6}, {12+\b*\b/\c/\c * (5-12)});

\coordinate (E) at ({\b*\b/\c/\c * 6 + \a*\b/\c/\c * (12-5)}, {12+\b*\b/\c/\c * (5-12) + \a*\b/\c/\c * 6});

\draw (A) -- (B) -- (C1) -- cycle;

\draw (A) -- (B) -- (C2) -- cycle;

\draw[->, thick] (A) -- node[right, xshift=6] {$b\cos\alpha\cdot\frac{\overrightarrow{AB}}{c}$} (D);

\draw[->, thick] (D) -- node[above, xshift=-18] {$b\sin\alpha\cdot\frac{\overrightarrow{AB}^\perp}{c}$} (E);

\path (A) node[above left] {A} -- (B) node[below right] {B} -- (C2) node[above right] {C} -- (C1) node[below left] {C'};

\path (A) -- node[below left] {c} (B) -- node

{a} (C2) -- node[above] {b} (A) -- (D) node[below left] {D};

\path (A) node[below right, xshift=10, yshift=-6] {$\alpha$};

\end{tikzpicture}

It follows that:

$$\overrightarrow C = \overrightarrow A + b\cos\alpha\cdot\frac{\overrightarrow{AB}}{c} \pm b\sin\alpha\cdot \frac{\overrightarrow{AB^\perp}}{c}

= \overrightarrow A + b\cdot \frac bc \cdot\frac{\overrightarrow{AB}}{c} \pm b\cdot \frac ac\cdot \frac{\overrightarrow{AB^\perp}}{c} \\

= \binom{0}{12} + \frac {68}{85} \cdot\binom{6-0}{5-12} \pm \frac {\sqrt{17\cdot 68}}{85}\cdot\binom{12-5}{6-0}

$$

\path (A) node[below right, xshift=10, yshift=-6] {$\alpha$};

\end{tikzpicture}

It follows that:

$$\overrightarrow C = \overrightarrow A + b\cos\alpha\cdot\frac{\overrightarrow{AB}}{c} \pm b\sin\alpha\cdot \frac{\overrightarrow{AB^\perp}}{c}

= \overrightarrow A + b\cdot \frac bc \cdot\frac{\overrightarrow{AB}}{c} \pm b\cdot \frac ac\cdot \frac{\overrightarrow{AB^\perp}}{c} \\

= \binom{0}{12} + \frac {68}{85} \cdot\binom{6-0}{5-12} \pm \frac {\sqrt{17\cdot 68}}{85}\cdot\binom{12-5}{6-0}

$$

- #5

Wilmer

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Nice. How about "without using trigonometry"? :)

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Wilmer said:Nice. How about "without using trigonometry"? :)

Trigonometry is only used in the first step.

We can replace it with non-trigonometry by observing that triangles ABC and ACD are

(Just added point D to my previous drawing for reference. ;))

- #7

Wilmer

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Ah so! Thanks loads.

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Wilmer said:Ah so! Thanks loads.

To be honest, I only used trigonometry because it means I have to use fewer words to explain those ratios! (Rofl)

- #9

Wilmer

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I'm also lazy!

- #10

tkhunny

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While you're at it, how about trying it without words, expressions, or symbols?

- #11

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tkhunny said:While you're at it, how about trying it without words, expressions, or symbols?

You mean just drawings? (Wondering)

Something like this:

\begin{tikzpicture}[>=stealth']

%preamble \usetikzlibrary{arrows}

\def\a{sqrt(17)}

\def\b{sqrt(68)}

\def\c{sqrt(85)}

\coordinate (A) at (0,12);

\coordinate (B) at (6,5);

\coordinate (C1) at (2,4);

\coordinate (C2) at ({38/5},{44/5});

\coordinate (D) at ({\b*\b/\c/\c * 6}, {12+\b*\b/\c/\c * (5-12)});

\coordinate (E) at ({\b*\b/\c/\c * 6 + \a*\b/\c/\c * (12-5)}, {12+\b*\b/\c/\c * (5-12) + \a*\b/\c/\c * 6});

\draw (A) -- (B) -- (C1) -- cycle;

\draw (A) -- (B) -- (C2) -- cycle;

\draw[->, thick] (A) -- (D);

\draw[->, thick] (D) -- (E);

\path (A) node[above left] {A} -- (B) node[below right] {B} -- (C2) node[above right] {C} -- (C1) node[below left] {C'};

\path (A) -- node[below left] {c} (B) -- node

{a} (C2) -- node[above] {b} (A);

\end{tikzpicture}

Admittedly there are still some symbols in there, but only the ones introduced in the OP!

The rest is left as an exercise to the reader. (Devil)

\end{tikzpicture}

Admittedly there are still some symbols in there, but only the ones introduced in the OP!

The rest is left as an exercise to the reader. (Devil)

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- #13

Wilmer

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GOOD idea...never thought of that!

- #14

tkhunny

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How are lines, arrows, and letters NOT symbols? They all mean things.

I just figured if we were eliminating things, we may as well eliminate it all. It's never been a favorite practice in my opinion - arbitrary elimination of reasonable methods. Anyway, exploration always has value.

- #15

Monoxdifly

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I like Serena said:For the record, something like that is called asangaku.

Doesn't

- #16

Wilmer

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Sangaku translation: calculation tablet) are Japanese geometrical problems or theorems on wooden tablets which were placed as offerings at Shinto shrines or Buddhist temples during the Edo period by members of all social classes

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Monoxdifly said:Doesn'tsangakusimply literally meanstriangle, though?

Google translates triangle as

And searching for sankakkei shows that this word is also used for a three dimensional puzzle of triangles.

View attachment 8074

However, that is different from a

View attachment 8075

- #18

Wilmer

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YIKES! I feel guilty for starting this thread :)

- #19

Monoxdifly

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I also feel guilty for posting an off-hand comment...

- #20

Wilmer

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x^2 + (y-12)^2 = 68

Expand:

x^2 - 12x + y^2 - 10y = -44

x^2 - 00x + y^2 - 24y = -76

Ahhh...ain't that cute!

2 solutions:

x=2, y=4 or x=38/5, y=44/5

The formula for finding the coordinates of the C point in a right triangle, given the coordinates of points A and B, is (x,y) = (xA + xB)/2, (yA + yB)/2. This means that the x-coordinate of C is the average of the x-coordinates of A and B, and the y-coordinate of C is the average of the y-coordinates of A and B.

The coordinates for a right triangle ABC are typically labeled as A(x_{1}, y_{1}), B(x_{2}, y_{2}), and C(x_{3}, y_{3}). The order in which the coordinates are listed does not matter, as long as they are correctly matched to their corresponding vertices.

No, the Pythagorean Theorem can only be used to find the length of the hypotenuse in a right triangle. It cannot be used to find the coordinates of any point in a triangle.

In order to find the coordinates of the C point in a right triangle, you will need at least two points and their coordinates. If you only have the length of the hypotenuse and one point's coordinates, you will not be able to accurately determine the coordinates of C.

Yes, if the right triangle is isosceles, meaning two of the sides have the same length, then the coordinates of C will be the midpoint of the line segment connecting the two equal sides. This is because an isosceles right triangle has a special property where the midpoint of the hypotenuse is also the midpoint of the triangle's altitude.

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