Solve Lesson 44 Page 82 SBT Math 10 – Kite>

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Topic

Given two straight linesfirst: mx – 2y – 1 = 0 and2: x – 2y + 3 = 0. With what value of parameter m then:

a)first //2?

b)first\( \bot {\Delta _2}\)?

Solution method – See details

Given 2 straight linesfirst: ax + by + c = 0 and2: a’x + b’y + c’ = 0. We havefirst //2 \( \Leftrightarrow \frac{a}{{a’}} = \frac{b}{{b’}} \ne \frac{c}{{c’}}\)

Step 1: Apply the above results to find m satisfiedfirst //2

Step 2: Find m to 2 VTPT offirst and2 multiplying scalars together by 0 satisfy ∆first\( \bot {\Delta _2}\)

Detailed explanation

first whose VTPT is \(\overrightarrow {{n_1}} = (m; – 2)\); ∆2 whose VTPT is \(\overrightarrow {{n_2}} = (1; – 2)\)

a)first //2 if and only if \(\overrightarrow {{n_1}} \) and \(\overrightarrow {{n_2}} \) are in the same direction and ∆first and2 not match

\( \Leftrightarrow \frac{m}{1} = \frac{{ – 2}}{{ – 2}} \ne \frac{{ – 1}}{3} \Leftrightarrow m = 1\)

So with m = 1 thenfirst //2

b) \({\Delta _1} \bot {\Delta _2} \Leftrightarrow \overrightarrow {{n_1}} .\overrightarrow {{n_2}} = 0 \Leftrightarrow m + 4 = 0 \Leftrightarrow m = – 4\)

So with m = -4 then \({\Delta _1} \bot {\Delta _2}\)

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