Question

The equation of the bisector of that angle between the lines $x+y=3$ and $2x-y=2$ which contains the point $(1,1)$ is

• A. $(\sqrt{5}-2\sqrt{2})x+(\sqrt{5}+\sqrt{2})y=3\sqrt{5}-2\sqrt{2}$
• B. $(\sqrt{5}+2\sqrt{2})x+(\sqrt{5}-\sqrt{2})y=3\sqrt{5}+2\sqrt{2}$
• C. $3x=2$
• D. none of these

Answer 1

The correct option is A $(\sqrt{5}-2\sqrt{2})x+(\sqrt{5}+\sqrt{2})y=3\sqrt{5}-2\sqrt{2}$

Equation of angle bisector of $x+y=3$ and $2x-y=2$ is
$\frac{x+y-3}{\sqrt{2}}=\pm \frac{2x-y-2}{\sqrt{5}}$
And for $(1,1)$ both eqaution are positive hence
required equation is
$\frac{x+y-3}{\sqrt{2}}=\frac{2x-y-2}{\sqrt{5}}$

$\sqrt{5}\times (x+y-3)=\sqrt{2}\times (2x-y-2)$

on rearranging the above equation,we get

$(\sqrt{5}-2\sqrt{2})x+(\sqrt{5}+\sqrt{2})y=3\sqrt{5}-2\sqrt{2}$