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Question 486311: Find the slope and equation of the line that contains the following 2 points A (3, 2) and (-6, 8)
Given the line y = (-3/5)x – 4 Find the equation of the line parallel to the given line and contains the point (-5 ,5)
Given the line y =( -2/5)x +3 Find the equation of the line perpendicular to the given line and contains the point (-4 ,-7)
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! x1 y1 x2 y2
3 2 -6 8
slope m = (y2-y1)/(x2-x1)
( 8 - 2 )/( -6 - 3 )
( 6 / -9 )
m= - 2/ 3
Plug value of the slope and point ( 3 , 2 ) in
Y = m x + b
2.00 = -2 + b
b= 2 - -2
b= 4
So the equation will be
Y = - 2/3 x + 4
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1 y = -3/5 x + -4
Divide by 1
y = -3/5 x -4
Compare this equation with y=mx+b
slope m = -3/5 0.8
The slope of a line parallel to the above line will be the same
The slope of the required line will be -3/5
m= -0.6 ,point ( -5 , 5 )
Find b by plugging the values of m & the point in
y=mx+b
5 = 3.00 + b
b= 2
m= -3/5
Plug value of the slope and b in y = mx +b
The required equation is y = -3/5 x + 2
y = - 3/ 5 x + 2
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y = -2/5 x -3
Compare this equation with y=mx+b
slope m = -2/5
The slope of a line perpendicular to the above line will be the negative reciprocal
m1*m2=-1
The slope of the required line will be 5/2
m= 5/2 ,point (-4,-7 )
Find b by plugging the values of m & the point in
y=mx+b
-7 = -10.00 + b
b= 3
m= 5/2
The required equation is y = 5/ 2 x + 3
m.ananth@hotmail.ca
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