Question

Tìm M để góc giữa hai đường d: ( m-3)x -(m-1)y +m-3=0 và ∆: ( m-2)x+(m+1)y-m-1 =0 , bằng 90°

Answer 1

Trước hết ta thấy \(\left(m-3\right)^2+\left(m-1\right)^2>0;\left(m-2\right)^2+\left(m+1\right)^2>0\forall m\)

Ta có: \(cos\left(d;\Delta\right)=cos90^o\)

\(\Leftrightarrow\dfrac{\left|\left(m-3\right)\left(m-1\right)-\left(m-1\right)\left(m+1\right)\right|}{\sqrt{\left(m-3\right)^2+\left(m-1\right)^2}.\sqrt{\left(m-2\right)^2+\left(m+1\right)^2}}=0\)

\(\Leftrightarrow\dfrac{\left|4-4m\right|}{\sqrt{\left(m-3\right)^2+\left(m-1\right)^2}.\sqrt{\left(m-2\right)^2+\left(m+1\right)^2}}=0\)

\(\Leftrightarrow m=1\)