Coordinate geometry help :( (1 Viewer)

JasonNg · Apr 2, 2012

If AB is a diameter of a circle with centre Q(4,5) and A = (8,3), find B.

Help please!

Carrotsticks · Apr 2, 2012

Q(4,5) and A(8,3)

To get from A to Q, we minus 4 from the x coordinate and add 2 to the y coordinate.

So similarly to get from Q to B, we do the exact same thing.

Doing so yields B(0,7)

There is a more precise algebraic way using gradients of collinear points, but I think this is the easiest.

JasonNg · Apr 2, 2012

Thanks Carrotsticks. Are you able to show me how to do this question?

Given the point A(7,8), find the coordinates of three points P with integer coordinates such that AP = root5.

Carrotsticks · Apr 2, 2012

JasonNg said:
Thanks Carrotsticks. Are you able to show me how to do this question?

Given the point A(7,8), find the coordinates of three points P with integer coordinates such that AP = root5.

This is a bit of a tricky question and there actually exists a total of 8 solutions.

$\\ PA=\sqrt{(x-7)^2+(y-8)^2} = \sqrt{5} \\\\ (x-7)^2+(y-8)^2 = 5 \\\\ $Now we must think 'What digits when squared and added... yield 5?' The answer (with a bit of thinking) would be $ \pm 1 $ and $ \pm 2. \\\\ $So for Case 1, we have $ x-7 = \pm 1 \Rightarrow x=6,8 $ and $ y-8 = \pm 2 \Rightarrow y=6,10 \\\\ $Now we must consider the case when it's the other way round ie: $ x=\pm 2 $ and $ y = \pm 1. \\\\ $Performing the same calculations and combining our answer yields the solutions...$

(6,6) (6,10) (8,6) (8,10) (5,7) (5,9) (9,7) (9,9)

Coordinate geometry help :( (1 Viewer)

JasonNg

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Carrotsticks

Retired

JasonNg

Member

Carrotsticks

Retired

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