Question 918075: find the equation of the line containing the point (-2,-5) and parallel to the line containing the points (-2,9) and (3,-10)
Answer by srinivas.g(540)   (Show Source):
You can put this solution on YOUR website! The point though which line is going be (x1,y1) = (-2,-5)
slope of a line parallel to a line passing through (x2,y2) and (x3,y3) is
= (y3-y2) /(x3-x2) , here lets say (x2,y2)= (-2,9) and (x3,y3) =(3,-10)
= =(-10-9)/(3-(-2))
=(-19)/(3+2)
=-19/5
equation of line passing though a point with given slope is given by
y-y1 =slope*(x-x1)
y-(-5) = (-19/5) *(x-(-2))
y+5 = (-19/5) *(x+2)
multiplying with 5 on both sides
5*(y+5)= 5* -(19/5) *(x+2)
5y+25 = -19*(x+2)
5y+25 = -19x-38
add 19x on both sides
19x+5y+25 =19x-19x-38
19x+ 5 y+25 =-38
add 38 on both sides
19x+5y+25+38 =-38+38
19x+5y+63 =0
So the required equation is 19x+ 5y+63 =0
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