Question 927703
On what axis? The x-axis? The y-axis? Never mind; we can find both.
 
Using the data given, a point-slope form of the equation of the line can be written as
{{{y-5=3(x-(-2))}}}<--->{{{y-5=3(x+2)}}}
A little algebra can be used to transform the equation above into the one and only slope-intercept form:
{{{y-5=3(x+2)}}}<--->{{{y-5=3(x+2)}}}<--->{{{y-5=3(x+2)}}}<--->{{{y-5=3(x+2)}}}
However, that is not necessary. The x- and y-intercepts can be found by making y=0 and x=0 respectively.
For that line,
the point on the x-axis, with {{{y=0}}} , has
{{{0-5=3(x+2)}}}<--->{{{-5=3(x+2)}}}<--->{{{-5/3=x+2}}}<--->{{{-5/3-2=x}}}<--->{{{x=-11/3}}} , and
the point on the y-axis, with {{{x=0}}} , has
{{{y-5=3(0+2)}}}<--->{{{y-5=3*2}}}<--->{{{y-5=6}}}<--->{{{y=6+5}}}<--->{{{y=11}}} .
So the other end-point of the segment, on a coordinate axis is either
{{{A(-11/3,0)}}} on the x-axis, or
{{{B(0,11)}}} on the y-axis.
{{{drawing(300,300,-8,2,-4,16,grid(0),
line(-5,-4,2,17),circle(-2,5,0.1),
circle(0,11,0.1),circle(-11/3,0,0.1),
locate(-6,2.6,A(-11/3,0)),locate(0.2,11.5,B(0,11)),
locate(-4,6,P(-2,5))
)}}}