find the equation of line passing through(2,-3) and perpendicular to the line 3x+4y-5
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3x+4y-5=0
y=(-3/4)x+(5/4)
m=(-3/4)
so the slope of the required line will be 4/3
that is it's equation will be
y=(4/3)x+c
now putting the co-ordinates of the given point in the equation...
(-3)=(4/3)(2)+c
c= -[(8/3)+3]
c= -17/9
so the equation of the required line will be...
y=(4/3)x-(17/9)
y=(-3/4)x+(5/4)
m=(-3/4)
so the slope of the required line will be 4/3
that is it's equation will be
y=(4/3)x+c
now putting the co-ordinates of the given point in the equation...
(-3)=(4/3)(2)+c
c= -[(8/3)+3]
c= -17/9
so the equation of the required line will be...
y=(4/3)x-(17/9)
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