Question

Answer 1

a, Gọi đt cần tìm là \(y=ax+b\left(a\ne0\right)\)

Theo đề: \(\left\{{}\begin{matrix}A\left(-2;-5\right)\inđths\\B\left(1;-4\right)\inđths\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2a+b=-5\\a+b=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=-\dfrac{13}{3}\end{matrix}\right.\)

Vậy đt cần tìm là \(y=\dfrac{1}{3}x-\dfrac{13}{3}\)

b, Gọi đt cần tìm là \(y=ax+b\left(a\ne0\right)\)

Theo đề: \(\left\{{}\begin{matrix}y=ax+b//y=5x-1\\M\left(3;-2\right)\inđths\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5\\b\ne-1\\3a+b=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5\\b=-17\end{matrix}\right.\)

Vậy đt cần tìm là \(y=5x-17\)