show that the points (4,5)(1,1)and (-2,5) are the vertices of an isoceles triangle. find the area?

Question

show your solution by step.draw a graph.

Answer 1

Every side of this triangle has a pair of ends and all the three pairs are cyclic. ie, if AB,BC & CA are the three sides of the given triangle , then the side AB has the ends (4,5),(1,1), BC has the ends (1,1)and (-2,5) & CA has the pair of ends (4,5),(-2,5).
By simple inspection of these pairs we can find that the points C and A are at the same height from the abcissa hence parallel to it. By this we can conclude easily that the length of CA=6 units.
Further, to be an isocelous triangle in this case the point B should be on the mid line of CA, and it is normal to CA. The mid point of CA is three units away from its ends C&A. As CA is parallel to absissa we can clearly conclude that any point between (1,0)&(1,5) will be the vertex of an isosceles triangle. Here the point B is (1,1) on the said line. Thus proved that the given points form an isosceles triangle.
We alredy know that CA = 6
And the other sides are equal to sqrt(4-1)^2 +(5-1)^2 = 5 units
Now, we can find the area of this triangle by Heron's Formula
ie sqrt[8(8-6)(8-5)^2] =12 sq units

Answer 2

I think you asked this question yesterday. If you still don't get it, you aren't even trying....

This is an example of 2 "3-4-5" triangles back to back forming an isosceles triangle with an area of 12. You should draw the points and dimension the lines to get the "big picture".