Linear Algebra 1 problem, Vector Geometry: Lines

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SUMMARY

The problem involves finding all points on the line L: x = (-3, 1) + t(1,-2) that are exactly 2 units away from the point (-3, 1). The solution yields the coordinates (3 ± 2 / √5, -1 ± 4/√5). To solve this, one can set up the equation of a circle centered at (-3, 1) with a radius of 2 and intersect it with the line equation. Alternatively, one can calculate the distance between the point (-3, 1) and points along the line defined by the parameter t until the distance equals 2.

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Given the line L: x = (-3, 1) + t(1,-2) find all x on L that lie 2 units from (-3, 1).
Problem: Given the line L: x = (-3, 1) + t(1,-2) find all x on L that lie 2 units from (-3, 1).

I know the answer is (3 ± 2 / √5, -1 ± 4/√5) but I don't know where to start. I found that if t=2, x= (-5, 5) and the normal vector is (2, 1) but I am not sure if this information is useful or how to use it.
 
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Given a choice of t, can you write down the distance between L(t) and (-3,1)?
 
Student323 said:
Summary:: Given the line L: x = (-3, 1) + t(1,-2) find all x on L that lie 2 units from (-3, 1).

Problem: Given the line L: x = (-3, 1) + t(1,-2) find all x on L that lie 2 units from (-3, 1).

I know the answer is (3 ± 2 / √5, -1 ± 4/√5) but I don't know where to start. I found that if t=2, x= (-5, 5) and the normal vector is (2, 1) but I am not sure if this information is useful or how to use it.
You could intersect the line with a circle around (-3,1) with radius 2. For that set up the equation for the circle and use it for (x,y) on the line.
 
Another way is to step along the line until you get to a point that is ##2## units apart. Calculate the length between ##\binom{-3}{1}## and ##\binom{-3}{1}+t\cdot \binom{1}{-2}## which equals ##2##.
 

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