MHB IBV16 Find a vector equation for the line

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The vector equation for the line is given as $\pmatrix{x \\ y} = \pmatrix{1 \\ 3} + t \pmatrix{5 \\ 2}$. This equation represents a line in two-dimensional space, where $\pmatrix{1 \\ 3}$ is a point on the line and $\pmatrix{5 \\ 2}$ indicates the direction of the line. There are infinite representations of this line, as any point on it combined with the direction vector can form a valid equation. The discussion clarifies that the x and y coordinates do not need to be different for the equation to be valid. Understanding this concept is crucial for accurately describing lines in vector form.
karush
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$\pmatrix{x \\ y} = \pmatrix{1 \\ 3} + t \pmatrix{5 \\ 2}$

this should be simple but still seem to miss them, also didn't know if the x,y should be different?:cool:
 
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What you have written looks good to me! :D

There are an infinite number of ways to write such a vector equation (as there are an infinite number of points on the line), but the one you chose is one of them. As long as you have a point on the line and the correct direction, you have described the line correctly.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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