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#