MHB *IBV6 Find vector eq of line passing thru (–1, 4),(3, –1). in form r = p + td

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The vector equation of the line passing through the points (–1, 4) and (3, –1) is expressed as r = p + td. The position vector p is defined as (3, -1), while the direction vector d is calculated as (4, -5). Thus, the final equation is r = (3, -1) + t(4, -5), where t is a real number. This formulation effectively represents the line in vector form. The solution is confirmed to be correct.
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Find a vector equation of the line passing through
$(–1, 4)$ and $(3, –1)$.
Give answer in the form $$r = p + td$$,
where $$t \in {R}$$

position vector would be $$(3, -1) = p$$

direction vector would be (3+1,-1-4) = (4,-5) = d

so $$r=(3,-1)+t(4,-5)$$
 
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