Gọi \(A\left(a;0\right),\left(B;b\right)\left(a,b>0\right)\)
Pt đường thẳng cần tìm có dạng :
\(\dfrac{x}{a}+\dfrac{y}{b}=1\)
Vì đường thẳng qua M(3;2) nên:
\(\dfrac{3}{a}+\dfrac{2}{b}=1\left(1\right)\)
a) \(0A+0B=12\Leftrightarrow a+b=12\Leftrightarrow a=12-b\left(2\right)\)
Thay (2) vào (1) ta có: \(\dfrac{3}{12-b}+\dfrac{2}{b}=1\)
\(\Leftrightarrow3b+2\left(12-b\right)=\left(12-b\right)b\)
\(\Leftrightarrow b^2-11b+24=0\Leftrightarrow b=3hayb=8\)
+ Với b=3=>a=9 => \(\dfrac{x}{9}+\dfrac{y}{3}=1\Leftrightarrow x+3y-9=0\)
+ Với b=8=>a=4 => \(\dfrac{x}{4}+\dfrac{y}{8}=1\Leftrightarrow2x+y-8=0\)
b) \(S_{\Lambda OAB}=\dfrac{1}{2}0A.0B=\dfrac{1}{2}ab=12\Leftrightarrow a=\dfrac{24}{b}\left(3\right)\)
Thay (3) vào (1) ta có: \(\dfrac{3b}{24}+\dfrac{2}{b}=1\Leftrightarrow b^2+16=8b\Leftrightarrow\left(b-4\right)^2=0\Leftrightarrow b=4\)
\(\Rightarrow a=6\Rightarrow\dfrac{x}{6}+\dfrac{y}{4}=1\Leftrightarrow2x+3y-12=0\)
