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Equation of a straight line passing through the point of intersection of \mathrm{x-y+1=0} and \mathrm{3 x+y-5=0} and perpendicular to one of them is

Option: 1

\mathrm{x+y+3=0}


Option: 2

\mathrm{x}+\mathrm{y}-3=0


Option: 3

\mathrm{x-3 y-5=0}


Option: 4

\mathrm{x+3 y+5=0}


Answers (1)

best_answer

Given lines are 3 \mathrm{x}+\mathrm{y}-5=0 \ldots (i)
and \mathrm{x-y+1=0 \ldots (ii)}

Slope line (i) is \mathrm{=-3} and slope of line (ii) is \mathrm{=1}

equation of any line through the point of intersection of the given lines is \mathrm{(3 x+y-5)+k(x-y +1)=0}
Since this line is perpendicular to one of the given lines,

\mathrm{\therefore \frac{3+\mathrm{k}}{\mathrm{k}-1}=-1\: or\: 1 / 3 \Rightarrow \mathrm{k}=-1 or -5}
Therefore, the required straight line is
\mathrm{\mathrm{x}+\mathrm{y}-3=0\, or \, \mathrm{x}-3 \mathrm{y}+5=0}

Posted by

shivangi.bhatnagar

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