What is an equation of the line that is perpendicular to y-4=2(x-6) and passes through the point (-3,-5)?
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y-4=2x-12
2x-y+4-12=0
2x-y-8=0
slope=-b/a
=-(-1/2)
=1/2
perpendicular slope= -2
passes to (-3,5)
equation is y-y1 =m(x-x1)
y-5 = -2(x+3)
y-5 = -2x-6
2x+y-5+6=0
2x+y+1=0 is answer obtained
2x-y+4-12=0
2x-y-8=0
slope=-b/a
=-(-1/2)
=1/2
perpendicular slope= -2
passes to (-3,5)
equation is y-y1 =m(x-x1)
y-5 = -2(x+3)
y-5 = -2x-6
2x+y-5+6=0
2x+y+1=0 is answer obtained
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