Find Point C Given Line L and Distance to B

[Telemachus]

Hi there. I'm tryin to find a point, let's call it C. I'm working on a R². What I know is that the point belongs to the line L: $$y=\displaystyle\frac{x}{2}+\displaystyle\frac{1}{2}$$ And that the distance to the point B(1,1), that belongs to L is $$\sqrt[ ]{20}$$.

How can I find it? I know there are two points, cause of the distance over the line.

I've tried to solve it using the distance pythagoric equation, but I don't know how to use the fact that B and C belongs to the same line.

 

[Mark44]

Hi there. I'm tryin to find a point, let's call it C. I'm working on a R². What I know is that the point belongs to the line L: $$y=\displaystyle\frac{x}{2}+\displaystyle\frac{1}{2}$$ And that the distance to the point B(1,1), that belongs to L is $$\sqrt[ ]{20}$$.

How can I find it? I know there are two points, cause of the distance over the line.

I've tried to solve it using the distance pythagoric equation, but I don't know how to use the fact that B and C belongs to the same line.

Every point on line L has coordinates (x, x/2 + 1/2). Set up an expression that represents the distance between this point and (1, 1), and set that expression to sqrt(20). Then solve the equation for x.

You should get two values for x, since there are two points on the line that are sqrt(20) units away from (1, 1).

 

[Telemachus]

Thanks Mark44. Heres my attempt to solve it:

$$\sqrt[ ]{20}=\sqrt[ ]{(1-x_0)^2+(1-y_0)^2}$$

$$(1-x_0)^2+(1-y_0)^2=20$$

So, I know that for any value of $$x_0$$, $$y_0$$ must be $$y_0=x_0/2 + 1/2$$ and I know $$(1-x_0)^2+(1-y_0)^2=20$$

Solving the system should I get the two values?

Bye there, and thanks again.

 

[Mark44]

Well, you can do that in one equation.
$$\sqrt{20}=\sqrt[ ]{(1-x_0)^2+(1-x_0/2 - 1/2)^2}$$

Yes, you should get two values for x₀.

 

[Telemachus]

Thank you.