Cho ba đường thẳng \({d_1}:3x-2y + 5 = 0\), \({d_2...

A

- Hướng dẫn giải

\(\left\{ \begin{array}{l} {d_1}:3x-2y + 5 = 0\\ {d_2}:2x + 4y-7 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x = - \frac{3}{8}\\ y = \frac{{31}}{{16}} \end{array} \right.\\ \to {d_1} \cap {d_2} = A\left( { - \frac{3}{8};\frac{{31}}{{16}}} \right).\)

Ta có:

\(\left\{ \begin{array}{l} A \in d\\ d||{d_3}:3x + 4y-1 = 0 \end{array} \right.\\ \to \left\{ \begin{array}{l} A \in d\\ d:3x + 4y + c = 0\,\,\left( {c\not = - 1} \right) \end{array} \right.\\ \to - \frac{9}{8} + \frac{{31}}{4} + c = 0 \Leftrightarrow c = - \frac{{53}}{8}.\)

Vậy \(d:3x + 4y-\frac{{53}}{8} = 0 \Leftrightarrow {d_3}:24x + 32y - 53 = 0.\)