MCQ
The distance between the lines $y = mx + c_1$ and $y = mx + c_2$ is:
- A$\frac{\text{c}_1-\text{c}_2}{\sqrt{\text{m}^2+1}}$
- B$\frac{|\text{c}_1-\text{c}_2|}{\sqrt{1+\text{m}^2}}$
- ✓$\frac{\text{c}_2-\text{c}_1}{\sqrt{1+\text{m}^2}}$
- D$0$
✓
Answer
Correct option: C.
$\frac{\text{c}_2-\text{c}_1}{\sqrt{1+\text{m}^2}}$
Let any point on the line $y = mx + c_1$ be $P(x_1, y_1).$
The equation of the other line is: $y = mx + c_2$
$\Rightarrow mx - y + c_2 = 0$
Distance of point $P$ from this line$, \text{d}=\frac{|\text{mx}_1-\text{y}_1+\text{c}_2|}{\sqrt{\text{m}^2+1}}$
Since $P$ line on the first line$,$ we get
$\Rightarrow y_1 = mx_1 + c_1$
$\Rightarrow mx_1 - y_1 = -c_1$
$\therefore \text{d}=\frac{|\text{c}_1-\text{c}_2|}{\sqrt{\text{m}^2+1}}$
The equation of the other line is: $y = mx + c_2$
$\Rightarrow mx - y + c_2 = 0$
Distance of point $P$ from this line$, \text{d}=\frac{|\text{mx}_1-\text{y}_1+\text{c}_2|}{\sqrt{\text{m}^2+1}}$
Since $P$ line on the first line$,$ we get
$\Rightarrow y_1 = mx_1 + c_1$
$\Rightarrow mx_1 - y_1 = -c_1$
$\therefore \text{d}=\frac{|\text{c}_1-\text{c}_2|}{\sqrt{\text{m}^2+1}}$
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