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If a line is perpendicular to the line $5 \mathrm{x}-\mathrm{y}=0$ and forms a triangle of area 5 square units with co-ordinate axes, then its equation is
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The correct answer is:
$\quad x+5 y \pm 5 \sqrt{2}=0$
$\quad 5 \mathrm{x}-\mathrm{y}=0$
$\mathrm{y}=5 \mathrm{x}$...(1)
Slope $=5$
Slope of perpendicular line will be $-\frac{1}{5}$
Let equation of line is $y=\left(-\frac{1}{5}\right) x+c$...(2)
Putting $\mathrm{y}=0$ $x=5 c$
$O B=5 c$
Intersecting point $\mathrm{A}$
Putting $\mathrm{x}=0$
$\mathrm{y}=-\frac{1}{5} \times 0+\mathrm{c}$
$\mathrm{y}=\mathrm{c}$
area $\Delta \mathrm{AOB}=\frac{1}{2} \times \mathrm{c} \times 5 \mathrm{c}$
$5=\frac{1}{2} \times 5 \mathrm{c}^{2}$
$c=\pm \sqrt{2}$
$\mathrm{y}=\left(-\frac{1}{5}\right) \mathrm{x} \pm \sqrt{2}$
$5 y+x \pm 5 \sqrt{2}=0$
$\mathrm{y}=5 \mathrm{x}$...(1)
Slope $=5$
Slope of perpendicular line will be $-\frac{1}{5}$
Let equation of line is $y=\left(-\frac{1}{5}\right) x+c$...(2)
Putting $\mathrm{y}=0$ $x=5 c$
$O B=5 c$
Intersecting point $\mathrm{A}$
Putting $\mathrm{x}=0$
$\mathrm{y}=-\frac{1}{5} \times 0+\mathrm{c}$
$\mathrm{y}=\mathrm{c}$
area $\Delta \mathrm{AOB}=\frac{1}{2} \times \mathrm{c} \times 5 \mathrm{c}$
$5=\frac{1}{2} \times 5 \mathrm{c}^{2}$
$c=\pm \sqrt{2}$
$\mathrm{y}=\left(-\frac{1}{5}\right) \mathrm{x} \pm \sqrt{2}$
$5 y+x \pm 5 \sqrt{2}=0$
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