Question

Find the equation of the line passing through the point of intersection of the lines $5x-8y+23=0$ and $7x+6y-71=0$ and perpendicular to the line $4x-2y=3$.

Answer 1

let the required line be $y=mx+c$

its perpendicular to $4x-2y=3$

Slope of $4x-2y=3$ $\Rightarrow$ $4x-2y=3\Rightarrow$ $m_1=\frac{4}{2}=2$

$m_1\times m=-1$

$2\times m=-1$

$m=-\frac{1}{2}x+c$

$\therefore$ $y=-\frac{1}{2}x+c$

Passes through intersection of $5x-8y+23=0$ and $7x+6y-71=0$

$3(5x-8y+23)=0\Rightarrow$ $15x-24y+69=0$

$64(7x+6y-71)=0$ $\Rightarrow$ $28x+24y-284=0$

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$43x=284-69$

$43x=215$

$x=\frac{215}{43}$

$x=5$

$5\left(5\right)-8y+23=0$

$8y=48\Rightarrow y=6$

meeting point $P=(5,6)$

$P$ passes through $y=mx+c$

$\Rightarrow$ $6=\frac{1}{2}\left(5\right)+c$

$c=6+\frac{5}{2}=\frac{17}{2}$

$\therefore$ $y=-\frac{x}{2}+\frac{17}{2}$

$\Rightarrow$ $2y+x-17=0$

Equation of the line is $2y+x-17=0$