Question 54979
write the equation of the line that passes through (3,2) and is perpendicular to the line 5x+4y=11
First, we need to find the slope of the line they gave us.  To do this we need to put the equation in slope intercept form {{{highlight(y=mx+b)}}}, where m=slope, and (0,b)=y-intercept.
{{{5x+4y=11}}}
{{{-5x+5x+4y=-5x+11}}}
{{{4y=-5x+11}}}
{{{4y/4=-5x/4+11/4}}}
{{{y=(-5/4)x+11/4}}}
{{{y=highlight(-5/4)x+11/4}}} the slope of this line is -5/4.
We need a line perpendicular to this line so our slope needs to be 4/5. (Flip the slope and change the sign.)
Now we use the point slope formula {{{highlight(y-y1=m(x-x1))}}}, where m=slope and (x1,y1)=the given point.  Our m=4/5 and our point is (x1,y1)=(3,2)
{{{y-2=(4/5)(x-3)}}}
{{{5(y-2)=5(4/5)(x-3)}}}
{{{5y-10=(20/5)(x-3)}}}
{{{5y-10=4(x-3)}}}
{{{5y-10=4x-12}}}
{{{5y-10+10=4x-12+10}}}
{{{5y=4x-2}}}
{{{5y/5=4x/5-2/5}}}
{{{y=(4/5)x-2/5}}} <--slope intercept form of a line {{{highlight(y=mx+b)}}}
Happy Calculating!!!