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P = ((2×12+3×2)/5, (2×8+3×3)/5) = (6,5).
P = ((1×8+4×(−2))/5, (1×(−6)+4×4)/5) = (0,2).
The trisection points divide AB in ratios 1:2 and 2:1, giving (4,3) and (7,5).
P = ((2×10+4)/3, (2×7−2)/3) = (8,4).
From the midpoint formula, 3=(1+x)/2 and 4=(2+y)/2, giving x=5 and y=6.
Using the section formula with P(4,5) gives m1:m2 = 1:2.
Since the point lies on the y-axis, its x-coordinate is zero. Applying the section formula gives the ratio 5:1.
AP = 3√2 and PB = 6√2, so AP:PB = 1:2.
y = (1×5 + 2×(−4))/3 = −1.
P = ((2×3−6)/3, (2×(−5)+4)/3) = (0,−2).
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