Question 498152: Hello!
Find the equation of the straight line,
trough (4,-2) and parallel with 2x+y-5 = 0.
Answer by Maths68(1474)   (Show Source):
You can put this solution on YOUR website!
We know that equation of the line
y=mx+b
2x+y-5 = 0 (Given)
y=-2x+5
m=-2 and b=5
Slope of the given line m = -2
Equation of the line passes through the point (4,-2)
Since both lines are parallel their slope will be same
m=-2 and Point (x1,y1)=(4,-2)
Now we have a point and slope, use the formula
increase in y / increase in x = slope
(y2-y1)/(x2-x1)=m
[(y-(-2)]/[(x-4]=-2
(y+2)/(x-4)=-2
y+2=-2(x-4)
y+2=-2x+8
y=-2x+8-2
y=-2x+6
y+2x-6=0 (Equation of the required line)
2nd Method
Put values of m and point (x,y) in the equation of the line
m=-2, (x,y)=(4,-2)
y=mx+b
-2=(-2)(4)+b
-2=-8+b
-2+8=b
b=6
Put the value of b and m in the equation of line, we have
y=mx+b
y=-2x+6
y+2x-6=0 (Equation of the required line)
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