એક સમબાજુતુષ્કોણની બે બાજુઓ રેખાઓ x - y + 1 = | vedclass

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Question

Equation of angle bisector of the lines $x-y+1=0$ and $7 \mathrm{x}-\mathrm{y}-5=0$ is given by

$\frac{x-y+1}{\sqrt{2}}=\pm \frac{7 x-y-5}{5 \sqrt{

Vedclass · Accepted Answer

$\left( {\frac{1}{3}, - \frac{8}{3}} \right)$

Vedclass · Answer

Equation of angle bisector of the lines $x-y+1=0$ and $7 \mathrm{x}-\mathrm{y}-5=0$ is given by

$\frac{x-y+1}{\sqrt{2}}=\pm \frac{7 x-y-5}{5 \sqrt{2}}$

$\Rightarrow 5(\mathrm{x}-\mathrm{y}+1)=7 \mathrm{x}-\mathrm{y}-5$

and

$5(x-y+1)=-7 x+y+5$

$	herefore 2 \mathrm{x}+4 \mathrm{y}-10=0 \Rightarrow \mathrm{x}+2 \mathrm{y}-5=0$ and

$12 x-6 y=0 \Rightarrow 2 x-y=0$

Now equation of diagonals are

$(x+1)+2(y+2)=0 \Rightarrow x+2 y+5=0$           .......$(1)$

and

$2(x+1)-(y+2)=0 \Rightarrow 2 x-y=0$           ....$(2)$

Clearly $\left(\frac{1}{3},-\frac{8}{3}ight)$ lies on $( 1)$

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