Question 1058171
Let's assume that the line intersects the x-axis at A:({{{a}}},{{{0}}}) and the y-axis at C:({{{0}}},{{{c}}}).
So you the distance is,
{{{(a-0)^2+(0-c)^2=(4sqrt(2))^2}}}
{{{a^2+c^2=32}}}
You also know that the midpoint is B and using the equation of the midpoint,
{{{x[b]=(a+0)/2=a/2}}}
and
{{{y[b]=(b+0)/2=c/2}}}
Since it's a common point between the line and the line {{{y=2x+2}}}, then,
{{{c/2=2(a/2)+2}}}
{{{c/2=a+2}}}
{{{c=2a+4}}}
Substituting above,
{{{a^2+(2a+4)^2=32}}}
{{{a^2+4a^2+16a+16=32}}}
{{{5a^2+16a-16=0}}}
{{{(a+4)(5a-4)=0}}}
Since you only allow solutions in Q1, only use the positive solution.
{{{5a-4=0}}}
{{{a=4/5}}}
then,
{{{c=2(4/5)+20/5}}}
{{{c=28/5}}}
You can then calculate the slope of the line,
{{{m=(c-0)/(0-a)=(28/5)/(-4/5)=-7}}}
So,
{{{highlight_green(y=-7x+28/5)}}}
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*[illustration lp17.JPG].