There are probably a lot of ways to solve this but here is the first method I thought of:
A quick sketch suggests the point (3,4) makes an angle of 90 degrees, so we want to prove that this is the case.
First label the angles at (3,4) A, (2,1) B and (-3,6) C, then label the opposite sides to these angles a,b and c for simplicity.
The Cosine rule states that for any angle in a triangle, Cos(A)=(b^{2}+c^{2}-a^{2})/2*b*c .
So we need to find the lengths of the three sides, which can be done using Pythagoras's theorem (a^{2}+b^{2}=c^{2})and the difference between the x co-ordinates and y co-ordinates for each point.
So a = distance between (2,1) and (-3,6) = sqrt{5^{2} +5^{2}} = sqrt{50},
b = distance between (3,4) and (-3,6) = sqrt{6^{2} +2^{2}} =sqrt{40},
c = distance between (3,4) and (2,1) = sqrt{1^{2} +3^{2}} =sqrt{10}.
Now Cos(A) = (b^{2}+c^{2}-a^{2})/2*b*c = (40 + 10 - 50)/40 = 0
Finally, the only possible value for A is 90 degrees in order for Cos(A)=0, and so the statement is proved!
Hope this helps.
-used sqrt{} to stand for 'the square root of' as math editor doesnt seem to work yet
Russell is right.
But you can also sort it out using Pythagoras Theorem.
First label the angles at (3,4) A, (2,1) B and (-3,6) C, then label the opposite sides to these angles a,b and c for simplicity.
Now find the length of each side (AB, BC && CA) using distance formula [ Distance=(sqrt(x1-x2)^{2 }+ (y1-y2)^{2}) ]
Now using pythagoras theorem : (hypotenuse)^{2 }= (base)^{2} + (perpendicular)^{2}^{}
If this is verified with its sides , it is proved that it is a right angled triangle.
given coordinates,(a1,b1) =(2,1)
(a2,b2)=(3,4)
(a3,b3)=(-3,6)
here we have one formula,area of a traingle with three coordinates=area=|a1(b2-b3)+a1(b3-b1)+a3(b1-b2)/2|
=|2(4-6)+2(6-1)+(-3)(1-4)/2|
=|2(-2)+2(5)+(-3)(-3)/2|
=|-4+10+9/2|
=|-15/2|=15/2