Question 56642
Find an equation of a line passing through (-1,1) and perpendicular to the line whose equation is given by 3x-2y=1.
First find the slope of the equation by putting it in slope intercept form:{{{highlight(y=mx+b)}}}, where m=slope and (0,b)=y-intercept:
{{{3x-2y=1}}}
{{{-3x+3x-2y=-3x+1}}}
{{{-2y=-3x+1}}}
{{{-2y/-2=-3x/-2+1/-2}}}
{{{y=(3/2)x-1/2}}}
{{{y=highlight((3/2))x-1/2}}}, the slope,m=3/2.
Perpendicular equations have slopes that are negative reciprocals of each other, so flip the slope over and change its sign.
We need a line with a slope, m=-2/3 going through the point (x1,y1)=(-1,1).
We need to use the point slope formula {{{highlight(y-y1=m(x-x1))}}}.
{{{y-1=(-2/3)(x-(-1))}}}
{{{y-1=(-2/3)(x+1)}}}
{{{y-1=(-2/3)x-(2/3)(1)}}}
{{{y-1=(-2/3)x-2/3}}}
{{{y-1+1=(-2/3)x-2/3+1}}}
{{{y=(-2/3)x-2/3+3/3}}}
{{{y=(-2/3)x+(-2+3)/3}}}
{{{y=(-2/3)x+1/3}}} <--Slope intercept form
{{{3y=-2x+1}}}
{{{2x+3y=1}}}  <--Standard form
Happy Calculating!!!