find the equation of the straight line perpendicular to the straight line x-2y+3=0 and passing through the point(1,-2)
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slope of the given line is
x-2y+3=0
-2y=-x-3
y = (1/2)x+(3/2)
so slope of this line is 1/2
by using this we can find the slope of line perpendicular to this line
m1 × m2=-1
1/2×m2=-1
m2 =-2
fir equation of line we use one point slope form
y-y1=m(x-x1)
we have x1=1 and y1=-2
on putting values in formula
y-(-2)=-2(x-1)
y+2=-2x+2
y=-2x
2x+y=0
hence the equation of line is 2x+y=o
x-2y+3=0
-2y=-x-3
y = (1/2)x+(3/2)
so slope of this line is 1/2
by using this we can find the slope of line perpendicular to this line
m1 × m2=-1
1/2×m2=-1
m2 =-2
fir equation of line we use one point slope form
y-y1=m(x-x1)
we have x1=1 and y1=-2
on putting values in formula
y-(-2)=-2(x-1)
y+2=-2x+2
y=-2x
2x+y=0
hence the equation of line is 2x+y=o
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