Line y =-x +6 has a point P(x,y) such that P is equidistant from A(10,-10) and Origin. PA = PO. Find P.
I found that PA = sqrt((10-x)^2 + (10-y)^2) = sqrt (200 +x ^2+y^2 +2y-2x) and PO = sqrt(x^2 + y^2)
Since PA = PO and squaring both sides I get
x^2 +y^2-2x+2y+200 = x^2 + y^2
x-y = 10
So using graph I think P in the ordered form is (8,-2) (where PA=PO = sqrt(68))
Is this how I find P... am I missing anything?
I see no errors in your work, and your reasoning seems sound at first glance. I think you, geoNewbie21, have solved this problem correctly.
Good job geoNewbie !!!
See the graph here : https://www.desmos.com/calculator/6wrak7ngnw
Notice that a cricle centered at (8, -2) with a radius of sqrt (68) will go through both the origin and (10-10)
So PA = PO = sqrt (68)
Thankyou Guest and CPhill,
CPhill could you actually share a link on how to use desmos.You have made it easy to understand/solve many geometry related questions, but I am not sure on how to use desmos to add the points A and O in this case or plot the circle :)
I don't really have a tuttorial on desmos
You can download it from their site.....you can log in by supplying an email and password so you can share your graphs (I believe these are "saved" so it makes logging in each time easy )
Just type whatever equation , points, etc., you want in the boxes on the left
If you check the label box under each entry it will display point coordinates.....you can also customize this by typing some identifier on the line by the label box (for ex : A = (1,0) ).....you can add anything on this label line
I like desmos OK, but I tend to use Geogebra more.....it is a little harder to use, but it is ideal for drawing segments, etc......you might explore it, as well