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. 5x + 2y – 4 = 0 B. 2x – 5y + 19 = 0 C. -5x + 2y – 16 = 0 D. 5x + 2y + 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: 5x + 2y + 4 = 0
Choose EASY
