Answer to Question #301157 in Geometry for Mazon

Question

ABC is a triangle and P, Q are the midpoints of AB, AC respectively. If AB = 2x

and AC = 2y, express the vectors (i) BC, (ii) PQ, (iii) PC, (iv) BQ in terms of x

and y. What can you deduce about the directed line-segments BC and PQ?

Answer 1

Solution.

$So, \space as \space 2x + BC = 2y, \newline BC = 2y - 2x. \newline As \space x + PQ = y, \newline PQ = y - x. \newline As \space x + PC = 2y, \newline PC = 2y - x. \newline As \space 2x + BQ = y, \newline BQ = y - 2x.$

Something we can deduce about the directed line-segments $BC$ and $PQ$ is that

$BC=2PQ$ , so they are parallel.

Answer:

$BC=2y−2x, \space PQ = y - x, \newline PC = 2y - x, \space BQ = y - 2x. \newline BC \space and \space PQ \space are \space parallel.$