By Melvin Hausner

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C) Orientation of P → Q = orientation of P′ → Q′. Under these conditions we say that P → Q determines the vector and that . Thus (read: vector PQ) is the symbol used for the vector determined by P → Q and the equation is a shorthand way of writing all three conditions (a), (b), and (c), above. It is necessary to discuss briefly the notion of parallelism and of orientation as used in (a) and (c) above. In Fig. 3, we certainly want . This suggests that PQ || P′Q′ [condition (a)] includes the case that line PQ is identical with line P′Q′.

In Fig. 31, find PT/TQ. 31 3. In Fig. 32, find BI/IC. 33 4. In Fig. 33, a, b, c, d, e, are given lengths. Find x. 5. In Fig. 34, C′ divides AB in the ratio 1/2, and similarly for A′ and B′. By assigning suitable masses to A, B, and C, and by suitably breaking them up, show that the centroid of A′B′C′ is also the centroid of ABC. Generalize and state a converse. 35 6. Let A, B, C, D be points in a plane. Prove that the three lines formed by joining the mid-point of any two of these points with the mid-point of the other two, all meet in a point.

Comparing (7) and (3), we obtain Equations (7) and (8) show that E is on segment BC and P is on segment AE as required. Finally, using Eq. (7), we obtain the answer to our problem: BE/EC = 5/4. The algebraically inclined will ask, “Why not solve for P as the unknown? ” This is done in the next example. 3 into an algebraic setting. 43 EXAMPLE 2. In Fig. 43, find P in terms of A, B, C and also find F in terms of A and C. Since we are in the land of algebra, we state the hypotheses in equation form and manipulate.