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**Topic**

Which of the following equations is a parametric equation for a line parallel to the line .?

*x* – 2*y* + 3 = 0?

A. \(\left\{ \begin{array}{l}x = – 1 + 2t\\y = 1 + t\end{array} \right.\) B. \(\left\{ \begin{ array}{l}x = 1 + 2t\\y = – 1 + t\end{array} \right.\) C. \(\left\{ \begin{array}{l}x = 1 + t\ \y = – 1 – 2t\end{array} \right.\)* *D. \(\left\{ \begin{array}{l}x = 1 – 2t\\y = – 1 + t\end{array} \right.\)

**Solution method – See details**

Step 1: Find the lines whose VTCP multiplies the scalar times the VTPT of the line *x* – 2*y* + 3 = 0 equals 0

Step 2: Get a point on the lines found in step 1, replace that point’s coordinates into the line PT

*x* – 2*y* + 3 = 0. If that point is not on the line *x* – 2*y* + 3 = 0, then the line containing that point is the line to find

**Detailed explanation**

Straight line ∆: *x* – 2*y* + 3 = 0 has a VTPT of \(\overrightarrow n = (1; – 2)\).

Straight line *d* parallel to ∆ take \(\overrightarrow n = (1; – 2)\) as VTPT and have VTCP as \(\overrightarrow u \) satisfying \(\overrightarrow u .\overrightarrow n = 0\)

(Type C, D)

Check points *USA*(-1; 1) belongs to the line \(\left\{ \begin{array}{l}x = – 1 + 2t\\y = 1 + t\end{array} \right.\). We see the coordinates *USA* satisfy PT *x* – 2*y* + 3 = 0 so M lies on (Type A)

** Select REMOVE**

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