Question

in triangle PQR,PQ=(w,x) andQR=(y,z) F is the mid point of the side PQ where w,x,y, and z are constants. (a) express RP and FR in terms of w,x,y and z (b) show that 2PQ +3QR+RP=2FR

Comments

The question seems to imply that (w,x) and (y,z) are coordinates, that is, references to two single points, not a line segment. A line segment (in this case the side of a triangle) would need a pair of coordinates to define it. Q is a common vertex to sides PQ and QR and could be defined by (w,x) or (y,z) but another coordinate would be needed to define P or R. Could you clarify the question either by editing it appropriately or adding a comment?

Answer 1

There are certain aspects to this question requiring clarification.

However, it's possible to produce a solution using three sets of coordinates for the vertices of PQR:

P(P_x,P_y), Q(Q_x,Q_y), R(R_x,R_y).

From these the midpoint of PQ is F(½(P_x+Q_x),½(P_y+Q_y).

(a)

Using Pythagoras' Theorem:

RP²=(R_x-P_x)²+(R_y-P_y)²; FR²=(½(P_x+Q_x)-R_x)²+(½(P_y+Q_y)-R_y)².

RP=√[(R_x-P_x)²+(R_y-P_y)²], FR=√[(½(P_x+Q_x)-R_x)²+(½(P_y+Q_y)-R_y)²].

(b)

PQ=√[(P_x-Q_x)²+(P_y-Q_y)²]; QR=√[(Q_x-R_x)²+(Q_y-R_y)²].

The proposition 2PQ+3QR+RP=2FR must apply generally to all triangles because no specific numerical lengths have been provided. So if we take a right triangle PQR such that PQ=10, QR=6 and PR=8, PF=FQ=5. 2PQ+3QR+RP=20+18+8=46, but by simple geometry, FR=5, so 2FR=10 and the proposition is clearly false. Since there is no given relation between w, x, y, z as independent variables, and no other constraints have been mentioned, some values assigned to the variables could produce the right triangle example, whatever these variables indicate. Nevertheless, we can derive relationships as follows:

Sine Rule: 

sinQP̂R/QR=sinPQ̂R/RP=sinQR̂P/PQ (triangle PQR);

sinQP̂R/FR=sinPF̂R/RP=sinFR̂P/PF (triangle PFR) (QP̂R and FP̂R are the same angle);

sinQF̂R/QR=sinPQ̂R/FR=sinQR̂F/FQ (triangle FQR) (PQ̂R and FQ̂R are the same angle).

And PF=FQ, FR̂P+QR̂F=QR̂P. Let a=PF=FQ; let b=QR̂P and c=FR̂P. (Lengths of sides and measures of angles.)

sinQR̂F=sin(b-c)=sin(b)cos(c)-cos(b)sin(c).

FR is common to triangles FQR and PFR, and PF=FQ=a, PQ=2a. The LHS of the proposed equation contains the green highlighted lengths. The other highlighting may help to understand the following logic.

More may follow in due course after clarification...