How do you write the vector equation that passes through point (3,-5) and parallel to <-2,5>?

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Answer 1 · 2016-11-03T13:43:42

$vecr=(3-2lambda)veci+5(lambda-1)vecj$

the vector equation of a line is given by;

$vecr=veca+lambdavecd$

where:

$veca$ is a known point on the line
$vecd$ is the direction of the line.

in this case:

$veca=<3,-5>$

$vecd=<-2,5>$

so we have $vecr=(3veci-5vecj)+lambda(-2veci+5vecj)$

$vecr=(3-2lambda)veci+5(lambda-1)vecj$