SATHEE: Straight Line Question 350

Straight Line Question 350

Question: Equation of a line passing through the point of intersection of lines $ 2x-3y+4=0, $ $ 3x+4y-5=0 $ and perpendicular to $ 6x-7y+3=0, $ then its equation is [RPET 2000]

Options:

A) $ 119x+102y+125=0 $

B) $ 119x+102y=125 $

C) $ 119x-102y=125 $

D)None of these

Show Answer

Answer:

Correct Answer: B

Solution:

The point of intersection of lines $ 2x-3y+4=0 $ and $ 3x+4y-5=0 $ is $ ( \frac{-2}{34},\frac{22}{17} ) $ The slope of required line $ =\frac{-7}{6} $ \ Equation of required line is $ y-\frac{22}{17}=\frac{-7}{6}( x+\frac{2}{34} ) $
Þ $ 119x+102y=125 $ .

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Straight Line Question 350

Question: Equation of a line passing through the point of intersection of lines $ 2x-3y+4=0, $ $ 3x+4y-5=0 $ and perpendicular to $ 6x-7y+3=0, $ then its equation is [RPET 2000]

Options:

Answer:

Solution:

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