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Where I think would be best to start with this question would be finding the length of AB, BC, and AC but I don't how I'd finish solving the question.

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    $\begingroup$ Compute the slopes of the three sides, and then determine if one is the negative reciprocal of another. $\endgroup$ – Paul May 23 '19 at 14:38
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    $\begingroup$ Or see if the lengths of the sides satisfy the Pythagorean Theorem. $\endgroup$ – paw88789 May 23 '19 at 14:39
  • $\begingroup$ @paw88789 After i find the side lengths, what should i do nxt $\endgroup$ – Vortext May 23 '19 at 14:42
  • $\begingroup$ Plot it! wolframalpha.com/input/?i=plot+%5B(2,6),(3,6),(9,0),(2,6)%5D $\endgroup$ – lhf May 23 '19 at 14:43
  • $\begingroup$ @Vortext: It's easier to find squares of side lengths and check whether the biggest number is equal to the sum of two smaller numbers. If it is, you have a right triangle. $\endgroup$ – Vasya May 23 '19 at 14:49
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Let the slope of two lines be $m_1$ and $m_2$. These two lines are perpendicular to each other if and only if:

$$m_1 \times m_2 = -1$$

Now compute the slopes of the lines and check if the above condition holds.

Another way, although slightly longer, way to do it would be to compute the sides of the triangle and checking for the Pythagoras theorem.

Edit: Also note this one other thing. If the slope of one of the lines comes out to be $0$, now what should one do, you may ask? That's a good question. Well then you know that if slope of a line is $0$ that means it is parallel to the x-axis. For another line to be perpendicular to it, that other line must be parallel to the y-axis.

Hope this helps.

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There are several possible approaches. You could use Pythagoras' theorem on the side lengths, or compute gradients and check if any two multiply to one another. Alternatively, you could compute vectors $AB, BC, CA$ and check if the dot product of any two is $0$.

In this case a geometric argument also works; note that $A$ and $C$ have the same $x$ coordinate. If the right angle is at $A$ or $C$, then $B$ must have the same $x$-coordinate as one of these, which is not true. If the right angle is at $B$, the $x$-coordinate of $B$ must be between that of $A$ and $C$ since it lies on the circle diameter $AC$. But this is also false.

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The squares of the distances are $$ AB^2 = 85, \quad AC^2=1, \quad BC^2=72 $$ The largest one is not the sum of the other two, and so the triangle does not satisfy the Pythagorean Theorem.

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Compute the pairwise dot products of the vectors $\overrightarrow{AB}$, $\overrightarrow{BC}$ and $\overrightarrow{CA}$. If any of them vanishes, then the angle between those sides of the triangle is a right angle.

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