Which of the following lines has no point of intersection with the line $y = 4 x + 5$ ?

y = \frac{1}{4} x - 4

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y = - \frac{1}{4} x - 5

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y = 4 x + \frac{1}{5}

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y = - 4 x + \frac{1}{5}

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y = - 4 x - \frac{1}{5}

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Solution

The correct option is C $y = 4 x + \frac{1}{5}$
Given, equation $y = 4 x + 5$ the slope of line is $4$ .
(A) $y = \frac{1}{4} x - 4$ the slope of line is $\frac{1}{4}$
(B) $y = - \frac{1}{4} x - 5$ the slope of line is $- \frac{1}{4}$
(C) $y = 4 x + \frac{1}{5}$ the slope of line is $4$
(D) $y = - 4 x + \frac{1}{5}$ the slope of line is $- 4$
(E) $y = - 4 x - \frac{1}{5}$ the slope of line is $- 4$
In these equation $A, B, D$ and $E$ the slope of line is different the given line the both line not parallel to given then both intersect the given line.
But in slope of equation (C) is equal to given line then both line parallal to each other then these not intersect each other.

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y = 4 x + 5

y

\frac{x + 3}{3} = \frac{y - 2}{- 2} = \frac{z + 1}{1}

4 x + 5 y + 3 z - 5 = 0

Find the angle between the following pair of lines

$\frac{x - 2}{2} = \frac{y - 1}{5} = \frac{z + 3}{- 3} a n d \frac{x + 2}{- 1} = \frac{y - 4}{8} = \frac{z - 5}{4}$

$\frac{x}{2} = \frac{y}{2} = \frac{z}{1} a n d \frac{z - 5}{4} = \frac{y - 2}{1} = \frac{z - 3}{8}$

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