$(d):y=4x$
a, $d_{2}:y=ax+b$
Vì $(d_{2})⊥d$
$⇒a=\dfrac{-1}{4}$
Gọi $(d_{2})∩Ox=C(0;-8)$
$⇒C∈(d_{2})$
$⇔-8=0a+b$
$⇔b=-8$
$⇒(d_{2}):y=-\dfrac{1}{4}a-8$
b, $(d_{3}):y=a'x+b'$
Vì $(d_{3})//(d)$
$⇒a'=4$
Gọi $(d_{3})∩Ox=A(0;b)⇒OA=|b|$
Gọi $(d_{3})∩Oy=B(\dfrac{-b}{4};0)⇒OB=|\dfrac{-b}{4}|$
Ta có: $S_{Δ_{AOB}}=\dfrac{1}{2}.AO.BO=\dfrac{1}{2}.|b|.|\dfrac{-b}{4}|$
$=\dfrac{b^{2}}{8}=8$
$⇔b^{2}=64$
$⇔b=±8$
$⇒(d_{3}):y=4x±8$
