**Topic**

Given the line \(\Delta :\left\{ \begin{array}{l}x = – 2 + 2t\\y = 3 – 5t\end{array} \right.\). Which of the following is a general equation for ∆?

A. 5*x* + 2*y* – 4 = 0 B. 2*x* – 5*y* + 19 = 0 C. -5*x* + 2*y* – 16 = 0 D. 5*x* + 2*y* + 4 = 0

**Solution method – See details**

Step 1: Find a normal vector of (take a vector multiply the scalar with VTCP of equal to 0)

Step 2: Find a passing point of (Get points in parametric PT)

Step 3: Find a general PT satisfying the above two conditions

**Detailed explanation**

∆ has a VTCP of \(\overrightarrow u = (2; – 5) \Rightarrow \) The VTPT of ∆ is \(\overrightarrow {{u_1}} = (5;2)\)or \(\overrightarrow {{u_2} } = ( – 5; – 2)\) Type B, C

∆ has a passing point of (-2; 3)

The PTTQ of is: 5*x* + 2*y* + 4 = 0 ** **

**Choose EASY**