Solve Lesson 28 Page 73 Math 10 – Kite >


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



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