Bài tập Viết phương trình đường thẳng đi qua 2 điểm lớp 12 (có lời giải)

A. \(\frac{{x + 4}}{1} = \frac{{y - 3}}{{ - 2}} = \frac{{z + 2}}{{ - 2}}\);

B. \(\frac{{x - 1}}{1} = \frac{{y - 1}}{{ - 2}} = \frac{{z - 2}}{{ - 2}}\);

C. \(\frac{{x + 1}}{{ - 5}} = \frac{{y + 1}}{2} = \frac{{z + 2}}{{ - 4}}\);

D. \(\frac{{x + 4}}{{ - 5}} = \frac{{y - 3}}{2} = \frac{{z + 2}}{{ - 4}}\).

A. \(\frac{{x - 1}}{1} = \frac{{y + 3}}{2} = \frac{{z - 7}}{2}\);

B. \(\frac{{x - 1}}{1} = \frac{{y + 3}}{2} = \frac{{z - 5}}{{ - 2}}\);

C. \(\frac{{x - 1}}{{ - 1}} = \frac{{y + 3}}{{ - 2}} = \frac{{z - 5}}{{ - 2}}\);

D. \(\frac{{x + 1}}{1} = \frac{{y - 3}}{2} = \frac{{z + 5}}{{ - 2}}\).

A. \(\left\{ \begin{array}{l}x = 1 + 2t\\y = 2t\\z = 1 + t\end{array} \right.\);

B. \(\left\{ \begin{array}{l}x = 1 + t\\y = t\\z = 1 + t\end{array} \right.\);

C. \(\left\{ \begin{array}{l}x = 1 - t\\y = t\\z = 1 + t\end{array} \right.\);

D. \(\left\{ \begin{array}{l}x = 1 + t\\y = t\\z = 1 - t\end{array} \right.\).

A. \(\left\{ \begin{array}{l}x = 1 + t\\y = - 2 + 2t\\z = - 1 - 3t\end{array} \right.\);

B. \(\left\{ \begin{array}{l}x = 1 + 3t\\y = - 2 - t\\z = - 3 + t\end{array} \right.\);

C. \(\left\{ \begin{array}{l}x = 1 + 2t\\y = 2 - 3t\\z = - 3 + 4t\end{array} \right.\);

D. \(\left\{ \begin{array}{l}x = - 1 + 2t\\y = - 2 - 3t\\z = 3 + 4t\end{array} \right.\).

A. \(\frac{{x - 1}}{1} = \frac{{y - 1}}{2} = \frac{{z + 1}}{3}\);

B. \(\frac{{x - 1}}{2} = \frac{{y - 1}}{3} = \frac{{z + 1}}{2}\);

C. \(\frac{{x + 2}}{1} = \frac{{y + 3}}{2} = \frac{{z + 2}}{3}\);

D. \(\frac{{x - 1}}{1} = \frac{{y - 2}}{1} = \frac{{z - 3}}{{ - 1}}\).

A. \(\frac{{x - 1}}{2} = \frac{{y + 2}}{3} = \frac{{z - 5}}{{ - 4}}\);

B. \(\frac{{x - 1}}{3} = \frac{{y + 2}}{1} = \frac{{z - 5}}{1}\);

C. \(\frac{{x + 1}}{2} = \frac{{y - 2}}{3} = \frac{{z + 5}}{{ - 4}}\);

D. \(\frac{{x - 3}}{1} = \frac{{y - 1}}{{ - 2}} = \frac{{z - 1}}{5}\).

A. \(\frac{{x + 3}}{1} = \frac{{y - 1}}{2} = \frac{{z + 2}}{{ - 2}}\);

B. \(\frac{{x - 3}}{1} = \frac{{y + 1}}{2} = \frac{{z - 2}}{{ - 2}}\);

C. \(\frac{{x + 1}}{3} = \frac{{y + 2}}{{ - 1}} = \frac{{z - 2}}{2}\);

D. \(\frac{{x - 1}}{3} = \frac{{y - 2}}{{ - 1}} = \frac{{z + 2}}{2}\).

A. \(\frac{{x + 1}}{4} = \frac{{y + 1}}{1} = \frac{{z - 1}}{1}\);

B. \(\frac{{x - 1}}{2} = \frac{{y - 1}}{{ - 1}} = \frac{{z + 1}}{3}\);

C. \(\frac{{x - 1}}{4} = \frac{{y - 1}}{1} = \frac{{z + 1}}{1}\);

D. \(\frac{{x + 1}}{2} = \frac{{y + 1}}{{ - 1}} = \frac{{z - 1}}{3}\).

A. \(\frac{{x - 1}}{3} = \frac{y}{1} = \frac{{z + 2}}{{ - 7}}\);

B. \(\frac{{x + 1}}{3} = \frac{y}{1} = \frac{{z - 2}}{{ - 7}}\);

C. \(\frac{{x - 1}}{1} = \frac{y}{1} = \frac{{z + 2}}{{ - 3}}\);

D. \(\frac{{x + 1}}{1} = \frac{y}{1} = \frac{{z - 2}}{3}\).

A. \(\frac{{x - 5}}{2} = \frac{{y - 4}}{1} = \frac{{z + 1}}{2}\);

B. \(\frac{{x + 1}}{4} = \frac{{y + 2}}{2} = \frac{{z + 3}}{{ - 4}}\);

C. \(\frac{{x - 1}}{4} = \frac{{y - 2}}{2} = \frac{{z - 3}}{4}\);

D. \(\frac{{x - 3}}{{ - 2}} = \frac{{y - 3}}{{ - 1}} = \frac{{z - 1}}{2}\).