## Solve Lesson 44 Page 82 SBT Math 10 – Kite>

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 $${\Delta _1} \bot {\Delta _2}$$