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Question 552954: determine the general equation of the line describe by the following
a)line passing through (-1,-3) and parallel to 2x-3y+6=0
b)perpendicular bisector of the segment form through (2,7)and (4,3)
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! 2 x -3 y = -6
Find the slope of this line
make y the subject
-3 y = -2 x + -6
Divide by -3
y = 2/ 3 x + 2
Compare this equation with y=mx+b
slope m = 2/ 3
The slope of a line parallel to the above line will be the same
The slope of the required line will be 2/ 3
m= 2/ 3 ,point ( -1 , -3 )
Find b by plugging the values of m & the point in
y=mx+b
-3 = - 2/ 3 + b
b= -7/ 3
m= 2/3
Plug value of the slope and b in y = mx +b
The required equation isy = (2/ 3)x-7/ 3
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(2,7)and (4,3)
use mid point formula
mid point x= (x+x2)/2, y= (y1+y2)/2
x=(4+2)/2=3
y=(3+7)/2 = 5
find the perpendicular line passing through (3,5) to 2x-3y=-6
2 x -3 y = -6
Find the slope of this line
-3 y = -2 x -6
Divide by -3
y = 2/ 3 x + 2
Compare this equation with y=mx+b
slope m = 2/3
The slope of a line perpendicular to the above line will be the negative reciprocal
m1*m2=-1
m= -1 1/2 ,point ( 3 , 5 )
Find b by plugging the values of m & the point in
y=mx+b
5 = -4.50 + b
b= 9.5
m= -3/2
The required equation is y = -3/ 2 x + 19/ 2
m.ananth@hotmail.ca
m.ananth@hotmail.ca
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