The two lines a1 x-b1 y-c1 and a2 x-b2 y-c2 are perpendicular ifa a1 a2-b1 b2-0b a1 b2-a2 b1c a1 b1-a2 b2-0d a1 b2-a2 b1-0

For two lines a1x + b1y = c1 nbsp; nbsp;and a2x + b2y = nbsp;c2 nbsp; nbsp; nbsp; Slope #160;is #160;given #160;by #160; #952;1=tan 1 b1a ...

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Mathematics

Equation of a Line Parallel to a Straight Line

The two lines...

Question

The two lines a₁x + b₁y = c₁ and a₂x + b₂y = c₂ are perpendicular if
(a) a₁a₂ + b₁b₂ = 0
(b) a₁b₂ + a₂b₁
(c) a₁b₁ + a₂b₂ = 0
(d) a₁b₂ + a₂b₁ = 0

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Solution

For two lines a₁x + b₁y = c₁and a₂x + b₂y = c₂
$Slope is given by (θ_{1}) = \tan^{- 1} (\frac{- b_{1}}{a_{1}}) i . e . {tanθ}_{1} = \frac{- b_{1}}{a_{1}} and {tanθ}_{2} = (\frac{- b_{2}}{a_{2}})$
We know, two lines are perpendicular if product of their slopes is –1.

$i . e . {tanθ}_{1} \times {tanθ}_{2} = - 1 i . e . \frac{- b_{1}}{a_{1}} \times \frac{- b_{2}}{a_{2}} = - 1 i . e . b_{1} b_{2} = - a_{1} a_{2} i . e . a_{1} a_{2} + b_{1} b_{2} = 0$

Hence, the correct answer is option A.

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The two lines a1x + b1y = c1 and a2x + b2y = c2 are perpendicular if (a) a1a2 + b1b2 = 0 (b) a1b2 + a2b1 (c) a1b1 + a2b2 = 0 (d) a1b2 + a2b1 = 0

The two lines a₁x + b₁y = c₁ and a₂x + b₂y = c₂ are perpendicular if
(a) a₁a₂ + b₁b₂ = 0
(b) a₁b₂ + a₂b₁
(c) a₁b₁ + a₂b₂ = 0
(d) a₁b₂ + a₂b₁ = 0