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vertices: A(-3, 5), B(4,7), and C(-1,-2)

Guest Aug 3, 2017
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4+0 Answers

 #1
avatar+1214 
+1

There are many ways to prove that a triangle is right, but this method, I believe, is the best and most efficient. Our first task is to find the actual length of the sides. To do this, I will utilize a fomula called the distance formula. The formula is the following:

 

\(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

 

Let's use this to find the distances:

 

\(d_{\overline{AB}}=\sqrt{(-3-4)^2+(5-7)^2}\)\(d_{\overline{BC}}=\sqrt{(4-(-1))^2+(7-(-2))^2}\)\(d_{\overline{CA}}=\sqrt{(-1-(-3))^2+(-2-5)^2}\)
\(d_{\overline{AB}}=\sqrt{(-7)^2+(-2)^2}\)\(d_{\overline{BC}}=\sqrt{5^2+9^2}\)\(d_{\overline{CA}}=\sqrt{2^2+(-7)^2}\)
\(d_{\overline{AB}}=\sqrt{49+4}=\sqrt{53}\)\(d_{\overline{BC}}=\sqrt{25+81}=\sqrt{106}\)\(d_{\overline{CA}}=\sqrt{4+49}=\sqrt{53}\)
   

 

Do you remember that famous formula called the Pythagorean theorem? It describes the relationship of the sides of a right triangle, stating that the sum of the squares of the legs is equal to the hypotenuse squared:

 

\(a^2+b^2=c^2\)

 

The converse happens to also be true; in other words, if \(a^2+b^2=c^2\), then the triangle is right. Let's check to see if this condition is true.

 

Before we do, however, we need to understand which sides are the legs and which is the hypotenuse. The hypotenuse is always the longest side of the triangle. Although we do not know the exact length of the segments AB, BC, and CA, we can determine which side is the longest since we know that \(AB=\sqrt{53}\)\(BC=\sqrt{106}\), and \(CA=\sqrt{53}\). Logically speaking,  \(\sqrt{106}>\sqrt{53}\) because the higher the radicand, the larger the actual value is. Now that we understand the hypotenuse is \(BC\), the legs do not matter. One is assigned the value and one is assigned the value b. To confirm that this triangle is indeed right, plug in the values into the equation:
 

\(a^2+b^2=c^2\)This is the Pythangorean theorem. Plug in the side lengths of the legs for a and b and the hypotenuse for c
\(\sqrt{53}^2+\sqrt{53}^2=\sqrt{106}^2\)Check to see if this condition is true. If it is, it validates that the triangle is indeed right. Of course, squaring a nonnegative number inside of a radical negates the radical.
\(53+53=106\) 
\(106=106\)This statement is true; 106=106. 
  

 

Because \(a^2+b^2=c^2\), we can conclude that this triangle is indeed right.

TheXSquaredFactor  Aug 3, 2017
edited by TheXSquaredFactor  Aug 3, 2017
 #2
avatar+18607 
+1

how do i prove triangleABC is a right triangle?

vertices: A(-3, 5), B(4,7), and C(-1,-2)

 

Let \(\vec{A} = \binom{-3}{5}\)
Let \(\vec{B} = \binom{4}{7}\)
Let \(\vec{C} = \binom{-1}{-2}\)

 

\(\begin{array}{rcll} \vec{CA} = \ ? \\ \vec{CA} &=& \vec{A} - \vec{C} \\ \vec{CA} &=& \binom{-3}{5} - \binom{-1}{-2} \\ \vec{CA} &=& \binom{-3+1}{5+2} \\ \vec{CA} &=& \binom{-2}{7} \\ \end{array}\)

 

\(\begin{array}{rcll} \vec{AB} = \ ? \\ \vec{AB} &=& \vec{B} - \vec{A} \\ \vec{AB} &=& \binom{4}{7} - \binom{-3}{5} \\ \vec{AB} &=& \binom{4+3}{7-5} \\ \vec{AB} &=& \binom{7}{2} \\ \end{array}\)

 

\(\triangle ABC\) is a right triangle if \(\vec{CA}\cdot \vec{AB} = 0\)

\(\begin{array}{rcll} \vec{CA}\cdot \vec{AB} = \ ? \\ \vec{CA}\cdot \vec{AB} &=& \binom{-2}{7} \cdot \binom{7}{2} \\ \vec{CA}\cdot \vec{AB} &=& (-2)\cdot 7 + 7 \cdot 2 \\ \vec{CA}\cdot \vec{AB} &=& -14+14 \\ \vec{CA}\cdot \vec{AB} &=& 0 \\ \end{array}\)

 

\(\mathbf{\triangle ABC}\) is a right triangle

 

laugh

heureka  Aug 3, 2017
 #3
avatar+71 
0

Heureka is absolutely right on this one - the question is testing your knowledge of vectors and the vector dot product.

It also shows you how vectors greatly simplify problems like this ,   X squared's solution is also right ,but look at the amount of work involved as opposed to using  vectors and the dot product.

 

ps  If you have the co-ordinates of any triangle's vertices you can of course find all 3 angles using the dot product. Always try and use it for problems like these as it will save you a lot of time and effort.

frasinscotland  Aug 3, 2017
 #4
avatar+76833 
+1

 

Actually, guest.....heureka's method is just different from that proposed by X2 ......if  the questioner has no knowledge of dot products and/or vectors, but understands the Pythagorean Theorem, then X2's  approach is good.....it just proves that there may be more than one way to solve something....and what is "best" depends upon your point of view......

 

 

cool cool cool

CPhill  Aug 3, 2017

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