SOLUTION: Find all points satisfying x+y=0 that are 8 units from (-2,3)

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Question 1205643: Find all points satisfying x+y=0 that are 8 units from (-2,3)
Found 2 solutions by josgarithmetic, math_tutor2020:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
x%2By=0
y=-x
The points or ordered pairs, (x,-x).

Question asks for all points (x,-x) which are 8 units distance from (-2,3).

sqrt%28%28x-%28-2%29%29%5E2%2B%28-x-3%29%5E2%29=8, applying the Distance Formula
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Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

Points that are 8 units from (-2,3) are on the circle %28x%2B2%29%5E2%2B%28y-3%29%5E2+=+64

Solve x+y = 0 for y to get y = -x

Plug y = -x into %28x%2B2%29%5E2%2B%28y-3%29%5E2+=+64 to get %28x%2B2%29%5E2%2B%28-x-3%29%5E2+=+64

Do a bit of algebra to get the following
%28x%2B2%29%5E2%2B%28-x-3%29%5E2+=+64

x%5E2%2B4x%2B4%2Bx%5E2%2B6x%2B9+=+64

x%5E2%2B4x%2B4%2Bx%5E2%2B6x%2B9-64+=+0

2x%5E2%2B10x-51+=+0

Now apply the quadratic formula with: a = 2, b = 10, c = -51

x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x+=+%28-10%2B-sqrt%28%2810%29%5E2-4%282%29%28-51%29%29%29%2F%282%282%29%29

x+=+%28-10%2B-sqrt%28100+%2B+408%29%29%2F%284%29

x+=+%28-10%2B-sqrt%28508%29%29%2F%284%29

x+=+%28-10%2B-++22.538855%29%2F%284%29

x+=+%28-10%2B22.538855%29%2F%284%29 or x+=+%28-10-22.538855%29%2F%284%29

x+=+%2812.538855%29%2F%284%29 or x+=+%28-32.538855%29%2F%284%29

x+=+3.134714 or x+=+-8.134714
Each decimal value is approximate.

Each of those x values is plugged into y = -x to find their corresponding paired y value.

The two answer points are approximately (3.134714, -3.134714) and (-8.134714, 8.134714) which I've marked as points A and B respectively.


Point A is located at roughly (3.134714, -3.134714)
Point B is located at roughly (-8.134714, 8.134714)
Point C is the center of the circle which is (-2,3)
AC = 8 and BC = 8 since any point on the circle is 8 units from point C.
Desmos and GeoGebra are two graphing tools I recommend.