The point (x,-1) is 13 units from the point (3,11), What are the possible values of x ?
Okay, to do all these problems relating to point coordinates. The first thing you do is to plot those points on the graph so you can have a clear picture of what it looks like. That's just a hint although you might be able to do this without plotting it, but it will be really helpful to do it.
Okay, so I gave you the distance formula on the previous question. Use it to find out what x2 is by substituting d = 13. You will end up with a sqroot of a number, which will give you TWO possible answers because the sqroot will give you a positive and a negative answer value (note that does not mean your x value have to be negative, I didn't calculate out the answer, but just let you know that x doesn't have to have one negative answer and one positive answer).
can you explain in details ,please
becouse I didn't understand
All right... so you have two points (x,-1) and (3,11) this is correlating to (x1,y1) and (x2,y2)in the distance formula.
d = sqroot [(x2-x1)^2 + (y2-y1)^2]
Plug in the numbers and you'll have:
13 = sqroot [(3-x)^2 + (11-(-1))^2]
13^2 = x^2-6x+9 + 12^2
169 = x^2-6x+153
0 = x^2-6x-16
0 = (x-8)(x+2)
Now you have two equations:
0 = (x-8) and 0 = (x+2)
Solving for x and you get x = 8 and x = -2 (this is what I meant by having two answers and one of them is negative, and my prediction was right ;p, I didn't work this problem out the first time).
So now you can answer your question. The two points that would give a distance of 13 units from point (3,11) are (-2,-1) and (8,-1).
If you are still doubting if you have the right answer, it's always a good thing to check it. So plot (3,11), (-2,-1) and (8,-1) and connect the two lines and see if they're roughly the same length. Okay so you're still not convinced, now look at the # of units that the two points are away from the x value of the (3,11). From -2 to 3, that's 5 units apart and from 8 to 3, it's also 5 units apart, so this suggest that they are very likely to have the same length.
I was hoping you can work out the problem by yourself, but it's cool that I worked it out for you, I haven't done math for like 3 years and I almost forgot how to factor out the x^2, lol!
I replied you, but I can't see it down here. Click on your first post of your question to view it.
why you write 13^2?
and from where you bring -6x?
can you explain?
sorry, I don't comprehend quickley
13^2 mean "13 square" or "13 x 13" or "13 raise to the power of 2". You have to square both sides of the equation to cancel off the "square root" on the right side of the equation so you can figure out "x", not "square root of x" because in the distance formula, x is contained inside the square root.
Where is the -6x comes from? Well, expand out the (3-x)^2. I just rewrite it as (-x+3)(-x+3), or you can keep it as (3-x)(3-x). I forgot the term, I think it's call FOILing or something like that, but basically you just multiply it like this:
(-x+3)(-x+3)
=(-x)(-x) + (-x)(3) + (3)(-x) + (3)(3)
= x^2 + (-3x) + (-3x) + 9
= x^2 - 6x + 9
and that's where the -6x comes from.
Now do you get how I got (x-8)(x+2)? This is the factoring step. Let me know if you don't get it.
ok 3^2 is 9 ?
why you put 6?
can you rewrite the solution please?
3^2 is where I get the +9 from... the -6x is from (-3x)+(-3x).
Solution = (-2,-1),(8,-1).