Question 306748
Suppose line l contains the points (5, 6) and (4, 3). Determine the equation of the line that: 
(1) Has the same y-intercept as l 
(2) Is perpendicular to l. 
<pre><font size = 3 color = "indigo"><b>
The other tutor's solution is a different problem:

Here are those points plotted:

{{{drawing(400,400,-10,10,-12,8, 
graph(400,400,-10,10,-12,8), 

line(5+.1,6,5-.1,6), line(5,6+.1,5,6-.1), line(5+.1,6+.1,5-.1,6-.1), line(5+.1,6-.1,5-.1,6+.1), locate(5,6,"(5,6)"), locate(4,3,"(4,3)"),

line(4+.1,3,4-.1,3), line(4,3+.1,4,3-.1), line(4+.1,3+.1,4-.1,3-.1), line(4+.1,3-.1,4-.1,3+.1) )}}}

Now we get a ruler and draw a green line through them:

{{{drawing(400,400,-10,10,-12,8, 
graph(400,400,-10,10,-12,8), 
green(line(14,33, -14,-51)),
line(5+.1,6,5-.1,6), line(5,6+.1,5,6-.1), line(5+.1,6+.1,5-.1,6-.1), line(5+.1,6-.1,5-.1,6+.1), locate(5,6,"(5,6)"), locate(4,3,"(4,3)"),

line(4+.1,3,4-.1,3), line(4,3+.1,4,3-.1), line(4+.1,3+.1,4-.1,3-.1), line(4+.1,3-.1,4-.1,3+.1) )}}}

We find the equation of that green line that contains the points
(5, 6) and (4, 3) by using the slope formula

{{{m=(y[2]-y[1])/(x[2]-x[1])=(3-6)/(4-5)=(-3)/(-1)=3}}}

Now we use the point-slope form:

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

{{{y-6=3(x-5)}}}

Now we simplify it to get the slope-y-intercept form

{{{y-6=3x-15}}}

{{{y=3x-9}}}

Comparing it to the slope-y-intercept form

{{{y = mx+b}}}, whose slope is m and whose y-intercept is (0,b)

We find that its y-intercept is (0,-9).

We can see in the graph above that the green line appears to 
have that y-intercept.

Now we want the equation of another line which is perpendicular
to that line.  It's slope will be the reciprocal of the slope 3
with the sign changed.  That is, the slope of the required line
will have slope {{{-1/3}}}.

This required line is to have the same y-intercept (0.-9), that
the given line has.

so its equation is

{{{y=-1/3}}}{{{x-9}}}

We already have one point on the required line, the y-intercept
(0,-9). We'll find another point on it, say, by substituting 
{{{x=-3}}}

{{{y=-1/3}}}{{{(-3)-9}}}
{{{y=""+1-9}}}
{{{y=-8}}}

So we see that the required line goes through (-3,-8) and (0,-9)

{{{drawing(400,400,-10,10,-12,8, 
graph(400,400,-10,10,-12,8), 
green(line(14,33, -14,-51)),
line(5+.1,6,5-.1,6), line(5,6+.1,5,6-.1), line(5+.1,6+.1,5-.1,6-.1), line(5+.1,6-.1,5-.1,6+.1), locate(5,6,"(5,6)"), locate(4,3,"(4,3)"),

line(4+.1,3,4-.1,3), line(4,3+.1,4,3-.1), line(4+.1,3+.1,4-.1,3-.1), line(4+.1,3-.1,4-.1,3+.1),

line(-3+.1,-8,-3-.1,-8), line(-3,-8+.1,-3,-8-.1), line(-3+.1,-8+.1,-3-.1,-8-.1), line(-3+.1,-8-.1,-3-.1,-8+.1), locate(-6,-8,"(-3,-8)")




 )}}}

Getting our ruler again, and drawing a blue line through (0,-9) and
(-3,-8) we have:

{{{drawing(400,400,-10,10,-12,8, 
graph(400,400,-10,10,-12,8), 
green(line(14,33, -14,-51)),
line(5+.1,6,5-.1,6), line(5,6+.1,5,6-.1), line(5+.1,6+.1,5-.1,6-.1), line(5+.1,6-.1,5-.1,6+.1), locate(5,6,"(5,6)"), locate(4,3,"(4,3)"),

line(4+.1,3,4-.1,3), line(4,3+.1,4,3-.1), line(4+.1,3+.1,4-.1,3-.1), line(4+.1,3-.1,4-.1,3+.1), locate(.4,-8.3,"(0,-9)"),

line(-3+.1,-8,-3-.1,-8), line(-3,-8+.1,-3,-8-.1), line(-3+.1,-8+.1,-3-.1,-8-.1), line(-3+.1,-8-.1,-3-.1,-8+.1), locate(-6,-8,"(-3,-8)"),
blue(line(-15,-4,12,-13))

 )}}}

The blue line looks very much perpendicular to the green line
and so we are satisfied that 

{{{y=-1/3}}}{{{x-9}}}

is the required equation of the required line. 

If you like you can put it in general form by multiplying through by 3,
then adding x to both sides:

{{{x+3y=-27}}}

Edwin</pre>