Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
Answer:
see attached
Step-by-step explanation:
We followed instructions to create the attached graph of the two triangles.
Each mid-segment is parallel to the side of the triangle whose midpoint is the vertex opposite that segment. So, we used a graphing tool to draw, for example, a line parallel to PQ through point R. The vertices of the original triangle are at the intersection points of these lines (or 1 mid-segment length from the midpoint).
__
Analytically, we can make use of the fact that each mid-segment is effectively a diagonal of a parallelogram. For example, PQ is a diagonal of APRQ. The other diagonal is AR. The two diagonals cross at the midpoint of each. This means we can find point A using the midpoint relation:
midpoint AR = midpoint PQ
(A +R)/2 = (P +Q)/2 . . . use the midpoint formula
A +R = P +Q . . . . . . . . multiply by 2
A = P +Q -R . . . . . . . . subtract R to get an expression for A
A = (2, 1) +(4, 5) -(7, 4) = (2+4-7, 1+5-4) = (-1, 2)
The other vertices of ΔABC can be found in similar fashion:
B = R +P -Q = (7, 4) +(2, 1) -(4, 5) = (7+2-4, 4+1-5) = (5, 0)
C = Q +R -P = (4, 5) +(7, 4) -(2, 1) = (4+7-2, 5+4-1) = (9, 8)
So, the original triangle's vertices are ...
A(-1, 2), B(5, 0), C(9, 8)
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.