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Is there a way to find the centre of enlargement without drawing rays?
I see once I have drawn rays, and found the centre, that there is a pattern in the co-ordinates of the respective shapes, related to the scale factor, but is there a kind of reverse process or similar; use the co-ordinates of the shape given, and the scale factor, to find the centre (without the rays)?
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The coordinates of two points before and after the enlargement is enough. Let's say they are (x1,y1) and (x2,y2) in the first shape and (X1,Y1) and (X2,Y2) in the transformed shape.
Find the equation of the line joining (x1,y1) to (X1,Y1) and also the other similar line. These are the rays but expressed algebraically.
Find where they cross. that's the centre.
You could probably construct a formula for this and then you're independent of a graph entirely.
Bob
If the centre is C and A = (x1,y1) and B = (X1,Y1) then CB/CA gives the scale factor.
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Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks, Bob.
"Find where they cross. that's the centre."
Can that be worked out from the two equations, without drawing the two rays?
I've worked out the following two equations from an example;
y=2x-2
y=(1/2)x-(1/2)
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Equating those values of y gives
In general
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks, Bob.
Why do we equate the two values of x?
I managed to follow your algebra for finding the value of x, thanks; great practice for me!
And the co-ordinates for the centre matched up with my graph, where the rays met (1,0), which was satisfying.
But why, once we established that x=1, do we conclude that y=0?
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Why do we equate the two values of x?
Actually I put the two ys equal and solved for x. But I could have eliminated the xs and solved for y. That would have worked as well.
If the lines cross then, at the intersection point they both have the same x and y coordinates. You have two equations so it becomes a simultaneous equations problem. What are the values of x and y that fit both equations simultaneously. So eliminate one unknown and solve for the other. As both equations are in the form y = function of x, the quickest way to an answer is to make the two functions of x equal and find the one x that works for both. Once you have that you can substitite that x value into either equation to get y. It works whichever equation you choose because that y is the one that fits in both equations.
Many routes to the same answer:
y = 2x -2
y = x/2 -1/2
Subtract the left hand sides and the right hand sides:
0 = 3x/2 - 3/2 so 3x/2 = 3/2 so x=1
substitute in y = 2x -2 ..... y = 2times 1 -2 = 0
substitute in y = x/2 -1/2 ...... y = 1/2 - 1/2 = 0
Make x the subject of each:
x = (y+2)/2
x = 2(y+1/2)
Set these equal
Substitiute in y = 2x - 2 ....... 0 = 2x - 2 ....... x = 1
Substitite in y = x/2 - 1/2 ....... 0 = x/2 - 1/2 ........ x/2 = 1/2 ....... x = 1
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks, Bob.
Prioritise. Persevere. No pain, no gain.
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