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Suppose I have a $8 \times 6$ rectangle, with its lower left corner at the origin $\left(0, 0\right)$. I want to scale this rectangle by $\frac{1}{2}$ at an anchor point $\left(3, 3\right)$. So the resulting rectangle is $4 \times 3$, but I cannot figure out how to compute the distance from the origin to the lower left corner of the new rectangle. Help is appreciated.
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1$\begingroup$ Sorry if I'm misunderstanding the construction, but doesn't that corner just go to the midpoint between the origin and (3,3)? In that case it's half the distance to (3,3), which is computed with the Pythagorean theorem. $\endgroup$– Jonas MeyerOct 1, 2010 at 4:19
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$\begingroup$ What do you mean by "anchor point"? Should the newly scaled rectangle have a corner at that point, or should it be centered there? $\endgroup$– J. M. ain't a mathematicianOct 1, 2010 at 4:20
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$\begingroup$ It is the midpoint for this case, but not for all cases. By anchor point I mean the center of scaling (as if the rectangle was moved to the origin and then scaled). $\endgroup$– MorrowlessOct 1, 2010 at 4:22
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$\begingroup$ Oh, I guess you're saying you really wanted a general way to find distances of transformed points. $\endgroup$– Jonas MeyerOct 1, 2010 at 4:32
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2 Answers
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Scaling by $0.5$ about the center $(3, 3)$ is the same as translating by $(-3, -3)$, scaling by $0.5$ about $(0, 0)$, then translating by $(3, 3)$. So, to find the coordinates of the lower left corner of the new rectangle, take the coordinates of the lower left corner of the original rectangle, subtract $3$ from each coordinate, multiply each coordinate by $0.5$, and add $3$ to each coordinate. Once you have the point's coordinates, you can find its distance from the origin.
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$\begingroup$ Thanks for the this, exactly what I needed and easy to picture. $\endgroup$ Oct 1, 2010 at 4:25
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$\begingroup$ It's just half the distance to the anchor point, right? The point started at the origin. $\endgroup$ Oct 1, 2010 at 4:27
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2$\begingroup$ @Jonas: I think so, in this case. I gave the more general explanation/method because often people ask what they think is a simpler question or what they ask ends up not being quite what they needed to know. $\endgroup$– IsaacOct 1, 2010 at 4:31
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1$\begingroup$ Well the real problem is that it's arbitrary. The rectangle, center of scale, and scale can be anything. But Isaac's method works for any case. $\endgroup$ Oct 1, 2010 at 4:37
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2$\begingroup$ If the scale is c and the anchor is (a,b), then the new point is (cx+(1-c)a,cy+(1-c)b), as easily follows from Isaac's method. $\endgroup$ Oct 1, 2010 at 4:41
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If you want to scale by .5 with anchor (a,b), then (x,y) will be sent to ((a+x)/2,((b+y)/2), the midpoint of the segment connecting the two points.
answered Oct 1, 2010 at 4:36