The point A has co-ordinates (5,4) and the point B has co-ordinates (x,7-x).
b) i) Expand (x-5)^2 <------- I did this by myself. Answer is x^2-10x+25.
That's the easy part, but I don't know these two:
ii) Show that AB^2= 2(x^2-8x+17).
iii) The minimum value of the curve x^2-8x+17 is 1.
Use this to find the minimum value of the distance AB as x varies.
I don't understand how to work these out, could you please show me how to do this and sort of explain the method if possible, but any help is great.
Thanks in advance.
b) i) Expand (x-5)^2 <------- I did this by myself. Answer is x^2-10x+25.
That's the easy part, but I don't know these two:
ii) Show that AB^2= 2(x^2-8x+17).
iii) The minimum value of the curve x^2-8x+17 is 1.
Use this to find the minimum value of the distance AB as x varies.
I don't understand how to work these out, could you please show me how to do this and sort of explain the method if possible, but any help is great.
Thanks in advance.
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I'm guessing that "AB²" means the square of the length of the line segment AB. If so, then
AB² = (x - 5)² + ((7 - x) - 4)² = x² - 10x + 25 + (3 - x)² = x² - 10x + 25 + 9 - 6x + x² =
= 2x² - 16x + 34 <----factor out a 2
=2(x² - 8x + 17).
That takes care of (i). Well, if the minimum of the curve y = x² - 8x + 17 is 1, and the square of the distance from A to B is 2(x² - 8x + 17), that would make the minimum of the squared distance equal to 2(1) = 2.
So the minimum distance is the square root of this---i.e. √(2).
AB² = (x - 5)² + ((7 - x) - 4)² = x² - 10x + 25 + (3 - x)² = x² - 10x + 25 + 9 - 6x + x² =
= 2x² - 16x + 34 <----factor out a 2
=2(x² - 8x + 17).
That takes care of (i). Well, if the minimum of the curve y = x² - 8x + 17 is 1, and the square of the distance from A to B is 2(x² - 8x + 17), that would make the minimum of the squared distance equal to 2(1) = 2.
So the minimum distance is the square root of this---i.e. √(2).