The distance between the lines y = mx + c1 and y = mx + c2 is:
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$\frac{\text{c}_1-\text{c}_2}{\sqrt{\text{m}^2+1}}$
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$\frac{|\text{c}_1-\text{c}_2|}{\sqrt{1+\text{m}^2}}$
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$\frac{\text{c}_2-\text{c}_1}{\sqrt{1+\text{m}^2}}$
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$0$
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M.C.Q (1 Marks)
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The distance between the lines y = mx + c1 and y = mx + c2 is:
$\frac{\text{c}_1-\text{c}_2}{\sqrt{\text{m}^2+1}}$
$\frac{|\text{c}_1-\text{c}_2|}{\sqrt{1+\text{m}^2}}$
$\frac{\text{c}_2-\text{c}_1}{\sqrt{1+\text{m}^2}}$
$0$
The equation of the straight line passing through the point (3, 2) and perpendicular to the line y = x is:
For specifying a straight line, how many geometrical parameters should be known?
The equations of the lines passing through the point (1, 0) and at a distance $\frac{\sqrt{3}}{2}$ from the origin, are
$\sqrt{3}\text{x}+\text{y}-\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}-\sqrt{3}=0$
$\sqrt{3}\text{x}+\text{y}+\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}+\sqrt{3}=0$
$\text{x}+\sqrt{3}\text{y}-\sqrt{3}=0,\text{x}-\sqrt{3}\text{y}-\sqrt{3}=0$
None of these.
One vertex of the equilateral triangle with centroid at the origin and one side as x + y - 2 = 0 is:
[Hint: Let ABC be the equilateral triangle with vertex A (h, k) and let $\text{D}(\alpha,\beta)$ be the point on BC. Then $\frac{2\alpha+\text{h}}{3}=0=\frac{2\beta+\text{k}}{3}.$ Also $\alpha+\beta-2=0$ and $\frac{\text{k}-0}{\text{h}-0}\times(-1)=-1\Big].$
| Column C1 | Column C2 | ||
| (a) | Parallel to y-axis is | (i) | $\lambda=-\frac{3}{4}$ |
| (b) | Perpendicular to 7x + y - 4 = 0 is | (ii) | $\lambda=-\frac{1}{3}$ |
| (c) | Passes through (1, 2) is | (iii) | $\lambda=-\frac{17}{41}$ |
| (d) | Parallel to x axis is | (iv) | $\lambda=3$ |
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