THIS IS AREA OF CIRCLES PLS ANSWER
2 answers:
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It's this graph. You didn't give me any options in the question.
Answer:
Check attached images.
Step-by-step explanation:
Oook, we're in for some number crunching.
Let's assume we already know that all 3 medians intersect in the same point A (same as the letter of the question, to make things easier), as do perpendicular bisectors of the sides in B, as do altitudes in C.*
a. Intersection of the medians is the easiest to compute, and its coordinates is the average of the vertex: hat's because it splits each medians in two parts, one being twice the other. By simple calculation, you can see the first point is A(1;3)
b. Intersection fof the right bisectors is done by picking two sides, writing down the equation of the bisectors, and watching where they meet . The easiest one to use is EG, since it's parallel to the x-axis, and the bisector goes through the middle point, red line in the image. As a second let's use the blue line. We know it goes to the midpoint of EF, which we can easily found with the average, and is perpendicular to EF (so the two slopes multiplied gives -1). Once we get the two lines we can simply calculate their intersect, which is the point .
c. Finally, for the altitudes. One is easy to spot, and it's the y axis, whose equation is x=0. The other, let's use the green line in my picture(s), since we know it's parallel to the blue line and saves us some computation: it passes through point G and has the same slope. Intersecting the two we get the third and last point, C(0;2).
d. there are plenty of ways to do so. if you know how to calculate the area of the triangle with determinants, you can plug the coordinates of A, B and C and see if you get 0.
Else, let's write the equation of the line between A and C, and see if B coordinates satisfies it.
That was quite the chore.
(*That two of them intersect is obvious, proof of the three theorems are about the third passing for the same point)
