P(−1,3),Q(7,−3)and R(4,1) are three points. Show that PQ=2QR using distance formula.
Answers
Step-by-step explanation:
Let a particle make an angle £ with the downward vertical while executing a vertical circle. Then its equation of motion is: T- mgcos£= mv^2/r, T= tension in the string, m= mass of the particle, v= speed, r= radius of the circle. Assuming that the bottom point of the circle is the zero plane of P.E., applying conservation of mechanical energy mv^2/2+mgr(1-cos£)= 0.5*m(v_b)^2 From this equation v^2= (v_b)^2–2gr(1-cos£). Using this equation one can find the velocity of a particle executing vertical circular motion.
Step-by-step explanation:
Given that PQR are three collinear points and
PQ=2.5. Let us now find the distance of PR
Recall Distance Formula is =$$ \sqrt { { \left( { x}_{ 2 }{ { x }_{ 1 } } \right) }^{ 2}+{ \left( { y }_{ 2 }{ { y }_{ 1 } } \right) }^{ 2 } } \\$$
Distance PR=
(113)
2
)+(104)
2
=
100
=10
Let us now consider S is the midpoint of PR,
then we have the co-ordinates of S as
2
11+3
,
2
10+4
=(7,7).
∴ it is clear that Q is the
midpoint of PS as we have know PQ=2.5 units.
Thus, to get the co-ordinates of Q take
midpoint of PS which is
2
3+7
,
2
4+7
=(5,11/2).