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Find the point C such that AC and BC form a 2:3 ratio
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(-1, 1.2) (-0.6, 3) (0, 2.4) (0.5, 2)
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@ganeshie8
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notice the 2 poins given, ( -3, 5) and (3, 0) use the distance equation to see how long the line AB is \(\large {\text{distance between 2 points}\\ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}}\)
ok... so you need \[\frac{AC}{BC} = \frac{2}{3}\] or 3AC = 2BC so the points are A(-3, 5) and B(3, 0) so you are looking at 3(-3, 5) and 2(3, 0) so for x its \[3 \times -3 + 2 \times 3 = \] then divide the answer by the sum of the parts in the ratio 2 + 3 same for y its \[3 \times 5 + 2 \times 0 =\] then divide the answer by the sum of the parts in the ratio.
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