Let the point be P, and the masses be located at P1 and P2.
The centre of mass is such that the moments of masses m1 and m2 exert an equal moment at the point P,
and the distance P1P2=d
namely,
m1(mPP1)=m2(mPP2)
The distance of the centre of mass from m1 is therefore
d1=d*m2/(m1+m2)
Similarly, the distance of the centre of mass from m2 is
d2=d*m1/(m1+m2)
Answer:
y = -x + 2
Step-by-step explanation:
Angle between these lines = 90°
Angle between BC and the angle bisector BD = 45°
Since, m(∠DBE) = 90° + 45° = 135°
Therefore, slope of the angle bisector BD = tan(90° + 45°)
= -tan(45)°
= -1
Let the equation of the angle bisector which passes through (x', y') and slope = m,
y - y' = m(x - x')
Where m = slope of the line = (-1)
Since, the angle bisector passes through a point B(-1, 3),
Equation of BD will be,
y - 3 = (-1)(x + 1)
y - 3 = -x - 1
y = -x - 1 + 3
y = -x + 2
Answer:
sorry I can't understand the answer sorry
