Question 172719
Given:....(2,0),{{{y=-5x+3}}}..
a)We know if it's parallel to the given Line {{{y=-5x+3}}}, they do have the same {{{slope=m=-5}}}
Therefore, thru points (2,0):
{{{0=-5*2+b}}}
{{{b=10}}}
So the eqn is ----->{{{highlight(y=-5x+10)}}}, Answer
As we see inthe graph;
{{{drawing(300,300,-10,10,-10,14,graph(300,300,-10,10,-10,14,-5x+3,-5x+10))}}}
---> RED>>>> {{{y=-5x+3}}}; GREEN>>>>{{{y=-5x+10}}}
b) Now it;s perpendicular:{{{m[2]=-1/m[1]}}}
So, {{{m[2]=-1/(-5)=1/5}}}
Then, thru points (2,0)
{{{0=(1/5)2+b}}}
{{{b=-2/5}}}
Then eqn ----------->{{{highlight(y=(1/5)x-(2/5))}}}
As we see the graph:
{{{drawing(300,300,-10,10,-10,14,graph(300,300,-10,10,-10,14,-5x+3,(1/5)x-(2/5))))}}} --->RED, {{{y=-5x+3}}}; GREEN, {{{(1/5)x-(2/5)}}}
Thank you,
Jojo