Question 167021
Let's get this line 1st: {{{3x – 4y = 9}}}
We can follow the eqn thru the SLope Intercept form, {{{y=mx+b}}}
{{{4y=3x-9}}} Divide whole eqn by 4
{{{cross(4)y/cross(4)=(3/4)x-(9/4)}}}---> slope (m)=3/4, and y-intercept(b) = -9/4
f(y)=0: Then,
{{{0=3x-9}}}
{{{3x=9}}} --> {{{cross(3)x/cross(3)=cross(9)3/cross(3)}}}
{{{x=3}}}
We have points (3,-9/4) for this line. We see the graph,
{{{drawing(300,300,-6,6,-6,6,grid(1),graph(300,300,-6,6,-6,6,(3/4)x-9/4),circle(3,0,.20),circle(0,-9/4,.20))}}} -- see points (3,-9/4)
Now, we'll see the line that passes thru (-2,-1) & perpendicular to above line:
By {{{y=mx+b}}}, and since it's perpendicular, the slope, {{{m[2]=-1/m[1]=-1/(3/4)=-4/3}}}
Continuing,
{{{-2=(-4/3)(-1)+b}}}
{{{-2=(4/3)+b}}}
{{{b=-2-4/3=(-6-4)/3}}}
{{{b=-10/3}}} ----> y-intercept
Therefore, getting this y intercept ( we'll need it in the graph of course), your line eqn ---> {{{y=(-4/3)x-(10/3)}}}
Simplifying if you want,
{{{highlight((4/3)x+y=-10/3)}}}, Or, Multiply whole eqn by 3:
{{{cross(3)(4/cross(3))x+3y=cross(3)(-10/cross(3))}}}
{{{highlight(4x+3y=-10)}}} ANSWER
See the graph,
{{{drawing(300,300,-5,5,-5,5,grid(1),graph(300,300,-5,5,-5,5,(-4/3)x-(10/3),(3/4)x-9/4),circle(-2,-1,.20),circle(3,0,.12),circle(0,-9/4,.12))}}} ---> see the RED line, it DOES NOT PASS on (-2,-1) BUT RATHER ON POINTS (-1,-2)as shown again:
{{{drawing(300,300,-5,5,-5,5,grid(1),graph(300,300,-5,5,-5,5,(-4/3)x-10/3,(3/4)x-9/4),red(circle(-1,-2,.20),circle(3,0,.12),circle(0,-9/4,.12)))}}} ---> RED Line {{{(-4/3)x-10/3}}}: GREEN Line {{{3x-4y=9}}}
Thank you,
Jojo