Find the equation of a line containing the point (-3,2) and perpendicular to the line y=(3/5)x+2

Find the equation of a line containing the point (-3,2) and perpendicular to the line y=(3/5)x+2
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1 Answer

strate  line thru (-3,2)

perpendikular tu slope=(3/5), so its  slope=-5/3

projekt from point tu x=0...deltax=3,

deltay=3*(-5/3)=-5

get point (0,-5)

y-intersept=-5

y=-(5/3) x -5
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