Question 166023
</pre><font size=4><b>
First we get the slope of line with points (2,3) & (1,5):
{{{m=(y[2]-y[1])/(x[2]-x[1])}}}
{{{m=(5-3)/(1-2)=2/(-1)}}}
{{{m=-2}}}, slope
Knowing the slop, we can get the intercept in point (1,1):
{{{y=mx+b}}}, SLOPE- INTERCEPT 
{{{1=-2(1)+b}}}
{{{b=1+2=3}}}
so, the eqn of the line: {{{highlight(y=-2x+3)}}} 
The line with points (2,3) & (1,5): same slope because  parallel, we'll check
VIA POINT SLOPE FORM:
{{{m=(y[2]-y[1])/(x[2]-x[1])}}}
{{{m=(5-3)/(1-2)=2/-1}}} ----> {{{m=-2}}}, same right?
For the intercept thru pints (2,3):
Via SLOPE INTERCEPT FORM, we pick points (2,3)
{{{3=-2(2)+b}}}
{{{b=3+4=7}}} ----> y intercept. The same value if you pick points (1,5).
Eqn of line: {{{y=-2x+7}}} (Green line) parallel to {{{y=-2x+3}}} (Red line)
See the graph,
{{{drawing(300,300,-7,7,-7,7,grid(1),graph(300,300,-7,7,-7,7,-2x+3,-2x+7),circle(1,1,.15),blue(circle(2,3,.15)),blue(circle(1,5,.15))))}}} --- see red line, it passes at point (1,1). Also green line passes points (2,3) & (1,5)
Thank you,
Jojo</pre>