Question 1078800
What is the equation of the line passing through (-3,2) with 
y-intercept two thirds of the x-intercept.
<pre><font size=4><b>
Let's draw an arbitrary line through (-3,2) that crosses the
y-axis at (0,b), where "b" is the y-intercept, and the x-axis
at (a,0) where "a" is the x-intercept.

{{{drawing(400,240,-5,5,-3,3,

line(0,-10,0,10), line(-10,0,10,0), circle(-3,2,.1), circle(0,1,.1), circle(3,0,.1), green(line(-12,5,12,-3)), locate(-3,2.4,"(-3,2)"),
locate(0,1.4,"(0,b)"), locate(3,.5,"(a,0)")  )}}}

Using the distance formula on (0,b) and (a,0) we find that
the slope is 

{{{m=(0-b)/(a-0)}}}{{{""=""}}}{{{-b/a}}}

Since we are told that the y-intercept "b" is two-thirds of the
x-intercept "a", we can write

{{{b}}}{{{""=""}}}{{{expr(2/3)a}}}

So the slope is {{{m}}}{{{""=""}}}{{{-b/a}}}{{{""=""}}}{{{-(expr(2/3)a)/a^""}}}{{{""=""}}}{{{-expr(2/3)a*expr(1/a)}}}{{{""=""}}}{{{-expr(2/3)cross(a)*expr(1/cross(a))}}}{{{""=""}}}{{{-2/3}}}

We use the point slope formula to find the equation
of the line through the point (-3,2) with slope {{{m}}}{{{""=""}}}{{{-2/3}}}:

{{{y-y[1]}}}{{{""=""}}}{{{m(x-x[1])}}}

{{{y-2}}}{{{""=""}}}{{{expr(-2/3)(x-(-3)^"")}}}

{{{y-2}}}{{{""=""}}}{{{expr(-2/3)(x+3^"")}}}

{{{y-2}}}{{{""=""}}}{{{expr(-2/3)x-2}}}

{{{y}}}{{{""=""}}}{{{expr(-2/3)x}}}

So the line goes through the origin and has this graph:
{{{drawing(400,240,-5,5,-3,3,
line(0,-10,0,10), line(-10,0,10,0), circle(0,0,.1), circle(0,0,.1), circle(-3,2,.1), green(line(-12,8,12,-8)), locate(-3,2.4,"(-3,2)"),
locate(0,-.3,"(a=0,0)"), locate(0,.5,"(0,b=0)")  )}}}

So we see that the x-intercept and the y-intercept are both zero,
and they coincide at the origin (0,0).  And, surprisingly, that's 
correct, since two-thirds of zero is indeed zero!  :)

Answer:  {{{y}}}{{{""=""}}}{{{expr(-2/3)x}}}

Edwin</pre></b>