Find the magnitude v and direction angle θ, to the nearest tenth of a degree, for the given vector v. v = -4i - 3j
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Answer:
<u><em>4 : </em></u> Coordinate of Q =
<em><u>5 : </u></em> Coordinate of point A = (-2,-2)
Explanation :
4) Given the points M(-3, -4) and T(5,0)
x₁= -3, y₁ = -4 and
x₂ = 5, y₂ = 0
m = 2, n = 1
Now apply the section formula,
{(mx₂+nx₁)/(m+n) , (my₂+ny₁)/(m+n)}
Coordinate of point Q =
5) AB:BC = 3:4
A = (x,y), end point
B = (4, 1)
C = (12, 5) end point
So, m = 3 & n = 4
x₁ = x , y₁ = y , x₂=12 & y₂ = 5
Apply section formula we get
4 = (36+4x)/7 & 1 = (15 + 4y)/7
28= 36 + 4x 7 = 4y + 15
4x =-8 4y =-8
x = -2 y = -2