Is there any other way of doing this problem.Below was the approach defined in Manhattan Quant Guide.
The horizontal distance between points A and B is 3 units (from -2 to -5).
Therefore, 4x = 3, and x = 0.75. The horizontal distance from B to the point is
x, or 0.75 units. The x-coordinate of the point is 0.75 away from -2, or
-2.75.
The vertical distance between points A and B is 6 units (from 0 to 6).
Therefore, 4x = 6, and x = 1.5. The vertical distance from B to the point is x,
or 1.5 units. The y-coordinate of the point is 1.5 away from 0, or 1.5.
The horizontal distance between points A and B is 3 units (from -2 to -5).
Therefore, 4x = 3, and x = 0.75. The horizontal distance from B to the point is
x, or 0.75 units. The x-coordinate of the point is 0.75 away from -2, or
-2.75.
The vertical distance between points A and B is 6 units (from 0 to 6).
Therefore, 4x = 6, and x = 1.5. The vertical distance from B to the point is x,
or 1.5 units. The y-coordinate of the point is 1.5 away from 0, or 1.5.
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