Find all the points having an x-coordinate of 3 whose distance from the point (-2, -1) is 13.

Question

anonymous · Answer

hey

anonymous · Answer

sorry trying to figure this out :)

skullpatrol · Answer

Start with a sketch please.

anonymous · Answer

ok

skullpatrol · Answer

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anonymous · Answer

I was told to use the distance formula and the 2 points are (3,y)

anonymous · Answer

wouldn't it be x -2 and y -1

anonymous · Answer

ok :)  d = √(x2-x1)^2+(y2-y1)^2 would I use my distance formula?

skullpatrol · Answer

Yes :) now just substitute the values in from your diagram.

anonymous · Answer

so it would be like this 13=√(x2-(-2))^2+(y2-(-1))^2  :)

skullpatrol · Answer

looks good

anonymous · Answer

13=x2+y2+3 then x is 3

skullpatrol · Answer

x2 = 3

anonymous · Answer

wow this is extremely difficult. wouldn't i be finding y  like this y =mx+b

anonymous · Answer

i already have X which is 3 right? so i would need the y point which is y = mx +b

saifoo.khan · Answer

13=(3)^2+y2+3
Solve for y.

anonymous · Answer

this is the distance formula as i had previous right?

phi · Answer

ok up to here
$$  13=\sqrt{(x2-(-2))^2+(y2-(-1))^2} $$ 
replace x2 with 3,  and just rename y2 to y (less typing)
$$ 13 = \sqrt{3 +2)^2 + (y+1)^2}$$
square both sides
169=  25 + (y+1)^2
can you finish ?

anonymous · Answer

got it :)

phi · Answer

you should get an ugly quadratic:
y^2 + 2y -143= 0
which you have to factor to find y

phi · Answer

another way to do this (assuming you are "allowed to"  is 
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