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Sagot :
Consider point P(x,y) such that P, X and Y are collinear,
As vectors
XP = XO + OZ where O(0,0)
XP = OZ - OX
XP= (x,y) - (-3,3)
XP = (x+3, y-3)
Similarly,
PY = (6-x, -3-y)
But XP= 2^PY
[x+3, y-3] = [2(6-x), 2(-3-y)]
Given both vectors are equal, as they go in the same direction, Solve for x and y accordingly:
x+3 = 12 - 2x
x = 3
y-3 = -6-2y
y = -1
Therefore, P(3,-1)
As vectors
XP = XO + OZ where O(0,0)
XP = OZ - OX
XP= (x,y) - (-3,3)
XP = (x+3, y-3)
Similarly,
PY = (6-x, -3-y)
But XP= 2^PY
[x+3, y-3] = [2(6-x), 2(-3-y)]
Given both vectors are equal, as they go in the same direction, Solve for x and y accordingly:
x+3 = 12 - 2x
x = 3
y-3 = -6-2y
y = -1
Therefore, P(3,-1)
The point P along the directed line segment from X(-3, 3) to Y(6,-3) that divides the segment in a 2:1 ratio is (3, -1).
What is section formula?
In coordinate geometry, Section formula is used to find the ratio in which a line segment is divided by a point internally or externally. It is used to find out the centroid, incenter and ex centers of a triangle
Given
[tex](x_{1} ,y_{1})[/tex] = X(-3, 3)
[tex](x_{2},y_{2} )[/tex] = Y(6,-3)
m: n = 2:1
Section formula = [tex](\frac{mx_{2}+nx_{1} }{m+n},\frac{my_{2}+ny_{1} }{m+n} )[/tex]
P(a, b) = [tex](\frac{2(6)+1(-3)}{2+1} ,\frac{2(-3)+1(3)}{2+1} )[/tex]
P(a, b) = [tex](\frac{12-3}{3},\frac{-6+3}{3} )[/tex]
P(a, b) = [tex](\frac{9}{3} ,\frac{-3}{3} )[/tex]
P(a, b) = (3, -1)
The point P along the directed line segment from X(-3, 3) to Y(6,-3) that divides the segment in a 2:1 ratio is (3, -1).
Find out more information about section formula here
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