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Question

Find the value of $$k$$ for which the lines $$x+2y+3=0$$ and  $$8x+ky-1=0$$ are parallel.


Solution

We have
x+2yy+3=02x−by+5=0
and, 8x+ky-1=ax+3y=2
Now,
x+2y+3=02x−by+5=0
or, 2y=-x-3by=2x+5
or $$y = \frac{{ - x}}{2} - \frac{3}{2}$$
Slope of the line  $$ = -\frac{2}{b}$$
Again
$$8x+ky-1=0$$
or $$ky=-8x+1$$
or $$y = \frac{{ - 8}}{R} + \frac{1}{R}$$
Solpe of the is line $$ = \frac{{ - 8}}{R}$$

$$\because $$ The lines are parallel, so the slopes of the two lines are equal 
$$\therefore \frac{{ - 1}}{2} = \frac{{ - 8}}{R}\,\,or\,\,R = 16$$
Hence, which is the required answer.

Mathematics

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