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Am I missing something here? My homework says to prove that the given triangle is a right triangle, but it does not appear to be a right triangle mathematically.

Let $A =(-3, 2)$, $B=(1, 0)$, and $C=(4,6)$.

Prove that $\blacktriangle ABC$ is a right triangle.

I have tried the slope method and it didn't work, I also tried the distance (Pythagorean) method and that didn't work either.

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  • $\begingroup$ Work out the squares of the side lengths and show $AB^2+BC^2=AC^2$ - no square roots required! $\endgroup$ – Mark Bennet Sep 23 '15 at 21:52
  • $\begingroup$ It works with both methods. Make a drawing first to see what angle is right. $\endgroup$ – A.Γ. Sep 23 '15 at 21:54
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Let's try the Pythagorean theorem:

Side $AB = [1- -3,0-2] = [4,-2]$, with length $\sqrt{20}$

Side $BC = [4- 1,6-0] = [3,6]$, with length $\sqrt{45}$

Side $AC = [4- -3,6-2] = [7,4]$, with length $\sqrt{65}$

$|AB|^2 + |BC|^2 = (\sqrt{20})^2+(\sqrt{45})^2 = 20 + 45 = 65 = (\sqrt{65})^2 = |AC|^2$

So the triangle must be right

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    $\begingroup$ I would call it by the name Law of cosines rather than Pythagorean theorem (which for me is just the other direction), but it is okay. $\endgroup$ – b00n heT Sep 23 '15 at 22:11
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I tried to solve it using the slope concept and indeed $A$, $B$ and $C$ are endpoints of a right triangle.

The slope of the line connecting $AB$ is $\frac{-1}{2}$.

The slope of the line connecting $AC$ is $\frac{4}{7}$.

Lastly, the slope of the line connecting $BC$ is 2.

Now since the product of the slope of line connecting $AB$ and the line connecting $BC$ is $\frac{-1}{2}\cdot 2=-1$, the line connecting $AB$ and the line connecting $BC$ are perpendicular by including line connecting $AC$ one can generate a triangle. This triangle a right triangle since the line connecting $AB$ and the line connecting $BC$ are perpendicular .

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Translate the triangle by $(-1,0)$. The new coordinates of the vertices become: $$ A(-4,2),\quad B(0,0),\quad C(3,6) $$ and since the dot product between $A$ and $C$ is zero, $\widehat{ABC}$ is a right angle.

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enter image description here

Add points $D=(-3,0)$ and $E=(4,0)$ and show that $\Delta ABD \sim\Delta BCE$. Then show that $\measuredangle ABC=90^\circ$.

Do you see why the slopes of perpendicular lines are negative reciprocals of each other?

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If you calculate again the distances you will see that $$AC^2=AB^2+BC^2$$

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