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Point of Intersection of Two Lines

The d.r's of ...

Question

The d.r's of the line $x = a y + b, z = c y + d$ are

1, a, c

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a, 1, c

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b, 1, c

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c, a, 1

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Solution

The correct option is B $a, 1, c$
Give lines,
$x = a y + b a n d z = c y + d$
$x = a y + b$
$⟹ a y = x - b$
$⟹ y = \frac{x - b}{a}$

Similarly from the other, $y = \frac{z - d}{c}$

$⟹ \frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c}$

D. R.'s of the given lines are $a, 1, c$

Suggest Corrections

Similar questions

Direction ratios of the line represented by the equation x = ay + b, z = cy + d are

x = a y + b, z = c y + d

x = a^{'} z + b^{'}, y = c^{'} z + d^{'}

x = a y + b, z = c y + d

\frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c}

x = a y + b, z = c y + d, x = a^{'} y + b^{'}, z = c^{'} y + d^{'}

a a^{'} + c c^{'}

a (y + z) = x, b (z + x) = y

c (y + z) = x,

a \neq - 1,

b \neq - 1,

c \neq - 1,

{(1 + a)}^{- 1} + {(1 + b)}^{- 1} + {(1 + c)}^{- 1}

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