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Finding a point on a line, given another point on the same line, and knowing its dist |
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| Jun8-10, 07:02 PM | #1 |
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Finding a point on a line, given another point on the same line, and knowing its dist
Hi there. I'm tryin to find a point, lets call it C. I'm working on a Rē. What I know is that the point belongs to the line L: [tex]y=\displaystyle\frac{x}{2}+\displaystyle\frac{1}{2}[/tex] And that the distance to the point B(1,1), that belongs to L is [tex]\sqrt[ ]{20}[/tex].
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. |
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| Jun8-10, 07:32 PM | #2 |
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You should get two values for x, since there are two points on the line that are sqrt(20) units away from (1, 1). |
| Jun8-10, 07:56 PM | #3 |
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Thanks Mark44. Heres my attempt to solve it:
[tex]\sqrt[ ]{20}=\sqrt[ ]{(1-x_0)^2+(1-y_0)^2}[/tex] [tex](1-x_0)^2+(1-y_0)^2=20[/tex] So, I know that for any value of [tex]x_0[/tex], [tex]y_0[/tex] must be [tex]y_0=x_0/2 + 1/2[/tex] and I know [tex](1-x_0)^2+(1-y_0)^2=20[/tex] Solving the system should I get the two values? Bye there, and thanks again. |
| Jun8-10, 08:19 PM | #4 |
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Finding a point on a line, given another point on the same line, and knowing its dist
Well, you can do that in one equation.
[tex]\sqrt{20}=\sqrt[ ]{(1-x_0)^2+(1-x_0/2 - 1/2)^2}[/tex] Yes, you should get two values for x0. |
| Jun9-10, 09:45 AM | #5 |
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Thank you.
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