# SOLUTION: This problem was similar to the other one I asked about. Perpendicular to the line y=2x-3;containing the point (1,-2) My answer was y=-1/2x and 1 1/2. My teacher remarked "use impr

Algebra ->  Algebra  -> Graphs -> SOLUTION: This problem was similar to the other one I asked about. Perpendicular to the line y=2x-3;containing the point (1,-2) My answer was y=-1/2x and 1 1/2. My teacher remarked "use impr      Log On

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 Algebra: Graphs, graphing equations and inequalities Solvers Lessons Answers archive Quiz In Depth

 Question 97226: This problem was similar to the other one I asked about. Perpendicular to the line y=2x-3;containing the point (1,-2) My answer was y=-1/2x and 1 1/2. My teacher remarked "use improper fractions" He wanted this graphed and two x y charts, mine were definitely wrong.Answer by edjones(7569)   (Show Source): You can put this solution on YOUR website!y=mx+b m=slope y=2x-3 m=2 b=-3 a line perpendicular to any line has slope of -1/m. Since m=2 then -1/m=-1/2=-.5 y-y[1]=m(x-x[1]) point slope form. For the perpendicular line: y-(-2)=-.5(x-1)) "containing the point (1,-2)" y+2=-.5(x-1) y+2=-.5x+.5 subtract 2 from both sides. y=-.5x-1.5 Ed Ed