# SOLUTION: Find an equation of the line containing the point (1,-2) and parallel to the line 9x-3y=5. Show work and explain each step. Please help!!!!!!!Don't not understand...Thanks

Algebra ->  Algebra  -> Inequalities -> SOLUTION: Find an equation of the line containing the point (1,-2) and parallel to the line 9x-3y=5. Show work and explain each step. Please help!!!!!!!Don't not understand...Thanks       Log On

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 Click here to see ALL problems on Inequalities Question 322079: Find an equation of the line containing the point (1,-2) and parallel to the line 9x-3y=5. Show work and explain each step. Please help!!!!!!!Don't not understand...Thanks Found 2 solutions by stanbon, Jk22:Answer by stanbon(57361)   (Show Source): You can put this solution on YOUR website! Find an equation of the line containing the point (1,-2) and parallel to the line 9x-3y=5. Show work and explain each step. --------------------- Given line: 9x-3y = 5 Slope-int form: y = 3x - (5/3) Note: slope = 3 ------ Find Equation of line with slope = 3, passing thru (1,-2) Form: y = mx + b Substitute: -2 = 3(1) = b b = -5 -------- Equation: y = 3x - 5 ============== Cheers, Stan H. ============== Answer by Jk22(389)   (Show Source): You can put this solution on YOUR website!A line is given by : set of (x,y)=OP+a*(dx,dy) OP is a point of the line, (dx,dy) is the vector giving the direction of the line. we need 2 points of the line : e.g. OP=(0,-5/3) and OP1=(1,4/3) hence : (dx,dy)=(1,3) --------------------- For the seeked line we put OP2=(1,-2) and the same vector of direction, to be parallel : line 2 is the set of :(x2,y2)=(1,-2)+a(1,3) x2=1+a y2=-2+3a eliminate a : 3x2-y2=5 hence : equation is : 9x-3y=15 (which is just first equation, with the RHS obtained by putting (x,y)=(1,-2) : 9-3*(-2)=15)