MCQ
For what value k, do the equations $3x - y + 8 = 0$ and $6x - ky + 16 = 0$ represent :
- A$\frac{1}{2}$
- B$-\frac{1}{2}$
- ✓$2$
- D$-2$
✓
Answer
Correct option: C.
$2$
Let $3x - y + 8 = 0 ...…(i)$
and $6x - (-ky) + 16 = 0 .....(ii)$
Here, $a_1=3, b_1=-1, c_1=8$
and $a_2=6, b_2=-k, c_2=16$
The given lines are coincident
$\therefore\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\therefore\frac{3}{6}=\frac{-1}{-\text{k}}=\frac{8}{16}$
$\therefore\frac{3}{6}=\frac{-1}{-\text{k}}$
$\therefore-\text{k}=-1\times\frac{6}{3}$
$\Rightarrow\text{k}=2$
and $6x - (-ky) + 16 = 0 .....(ii)$
Here, $a_1=3, b_1=-1, c_1=8$
and $a_2=6, b_2=-k, c_2=16$
The given lines are coincident
$\therefore\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\therefore\frac{3}{6}=\frac{-1}{-\text{k}}=\frac{8}{16}$
$\therefore\frac{3}{6}=\frac{-1}{-\text{k}}$
$\therefore-\text{k}=-1\times\frac{6}{3}$
$\Rightarrow\text{k}=2$
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