A + (7/10)(B -A) = (7/10)B + (3/10)A
... = (7B + 3A)/10 = (7(11, 4) +3(-4,-6))/10
... = (77 -12, 28 -18)/10
... = (6.5, 1)
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The way this is done using geometric construction is to use similar triangles. Here, the line segment AC is constructed to be 10 units long, and AD is made to be 7 units. Then the line DE parallel to CB will intersect AB at the desired point (E). Point E has the coordinates computed above.
