Cho hai đường thẳng: (d1): a1x + b1y + c1 = 0 (a1, b1 khác 0); (d2): a2x + b2y + c2 = 0

* Hướng dẫn giải

Như ta đã học ở phần trước, với hai đường thẳng $\left({\mathrm{d}}_1\right):\mathrm{y}={\mathrm{m}}_1\mathrm{x}+{\mathrm{n}}_1$ và $\left({\mathrm{d}}_2\right):\mathrm{y}={\mathrm{m}}_2\mathrm{x}+{\mathrm{n}}_2$

Ta có các kết quả:

- $\left({\mathrm{d}}_1\right)//\left({\mathrm{d}}_2\right)\iff {\mathrm{m}}_1={\mathrm{m}}_2,{\mathrm{n}}_1\ne {\mathrm{n}}_2$

- $\left({\mathrm{d}}_1\right)\equiv \left({\mathrm{d}}_2\right)\iff {\mathrm{m}}_1={\mathrm{m}}_2,{\mathrm{n}}_1={\mathrm{n}}_2$

- $\left({\mathrm{d}}_1\right)\cap \left({\mathrm{d}}_2\right)\iff {\mathrm{m}}_1\ne {\mathrm{m}}_2$

Viết lại các đường thẳng dưới dạng: $\left({\mathrm{d}}_1\right):\mathrm{y}=-\frac{{\mathrm{a}}_1}{{\mathrm{b}}_1}\mathrm{x}+\frac{{\mathrm{c}}_1}{{\mathrm{b}}_1}$ và $\left({\mathrm{d}}_2\right):\mathrm{y}=-\frac{{\mathrm{a}}_2}{{\mathrm{b}}_2}\mathrm{x}+\frac{{\mathrm{c}}_2}{{\mathrm{b}}_2}$

1. ${\mathrm{d}}_1,{\mathrm{d}}_2$ cắt nhau khi $-\frac{{\mathrm{a}}_1}{{\mathrm{b}}_1}\ne -\frac{{\mathrm{a}}_2}{{\mathrm{b}}_2}\iff \frac{{\mathrm{a}}_1}{{\mathrm{b}}_1}\ne \frac{{\mathrm{a}}_2}{{\mathrm{b}}_2}$

2. ${\mathrm{d}}_1,{\mathrm{d}}_2$ song song với nhau khi

$$\begin{gathered} \left\{\begin{array}{c}-\frac{{\mathrm{a}}_1}{{\mathrm{b}}_1}=-\frac{{\mathrm{a}}_2}{{\mathrm{b}}_2}\\ \frac{{\mathrm{c}}_1}{{\mathrm{b}}_1}\ne \frac{{\mathrm{c}}_2}{{\mathrm{b}}_2}\end{array}\right.\iff \left\{\begin{array}{c}\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\\ \frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\ne \frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}\end{array}\right. \\ \iff \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\ne \frac{{\mathrm{c}}_1}{{\mathrm{c}}_2} \end{gathered}$$

3. ${\mathrm{d}}_1,{\mathrm{d}}_2$ trùng nhau khi

$$\begin{gathered} \left\{\begin{array}{c}-\frac{{\mathrm{a}}_1}{{\mathrm{b}}_1}=-\frac{{\mathrm{a}}_2}{{\mathrm{b}}_2}\\ \frac{{\mathrm{c}}_1}{{\mathrm{b}}_1}=\frac{{\mathrm{c}}_2}{{\mathrm{b}}_2}\end{array}\right.\iff \left\{\begin{array}{c}\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\\ \frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}\end{array}\right. \\ \iff \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2} \end{gathered}$$