Question 192965
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Given Line: {{{2x+y=7}}}{{{red(EQN1)}}} ---> {{{y=highlight(-2)x+7}}}, where, 
{{{Slope=m=-2}}}
Therefore, the line passing thru (4,4) that is perpendicular has a Slope, {{{m[2]=-1/m=-1/-2=red(1/2)}}}.


But remember the intercepts of {{{red(EQN1)}}}:
Let Fy=0:
{{{2x+0=7}}} ---> {{{cross(2)x/cross(2)=7/2}}}
{{{red(x=7/2)}}}, or (3.5,0), X-Intercept of {{{red(EQN1)}}}
Let Fx=0:
{{{2(0)+y=7}}}
{{{red(y=7)}}}, Y-Intercept of {{{red(EQN1)}}}



On {{{red(EQN2)}}}:
Thru point (4,4),
{{{y=mx+b}}}
{{{4=(1/cross(2))(cross(4)2)+b}}}
{{{b=4-2=2}}}, Y-Intercept


It follows, {{{red(y=(1/2)x+2)}}} {{{red(EQN2)}}}(Answer)-->  Line passing thru (4,4)


Let Fy=0:
{{{0=(1/2)x+2}}}
{{{-2=(1/2)x}}}
{{{x=2*-2=-4}}}, X-Intercept


We see the line;
{{{drawing(400,400,-7,7,-4,10,graph(400,400,-7,7,-4,10,-2x+7,(1/2)x+2),blue(circle(-4,0,.12)),blue(circle(0,2,.12)),blue(circle(4,4,.12)),blue(circle(3.5,0,.12)),blue(circle(0,7,.12)))}}} ----> {{{red(RED=EQN1)}}}; {{{green(GREEN=EQN2)}}}


Thank you,
Jojo</font>