Question

Lines $L_1\equiv y-x=0$ and $L_2\equiv 2x+y=0$ intersect the line $L_3:y+2=0$ at $P$ and $Q$ respectively. The bisector of the acute angle between $L_1$ and $L_2$ intersects $L_3$ at $R$.

STATEMENT -1 : The ratio $PR:RQ=2\sqrt{2}:\sqrt{5}.$

STATEMENT-2: In any triangle, bisector of an angle divides the triangle into two similar triangles.

• A. Statement-1 is True, Statement -2 is True; Statement -2 is a correct explanation for Statement-1
• B. Statement -1 is True, Statement -2 is True; Statement -2 is not a correct explanation for Statement -1
• C. Statement-1 is True, Statement -2 is False
• D. Statement -1 is False, Statement -2 is True

Answer 1

The correct option is C Statement-1 is True, Statement -2 is False

Given

$L_1\equiv y-x=0$-----(1)

$L_2\equiv 2x+y=0$--------(2)

$L_3\equiv y+2=0$------(3)

P is the intersecting point of line $L_1$ and $L_3$

on solving eq (1) and (3) we get

$P(-2,-2)$

Q is the intersecting point of line $L_2$ and $L_3$

on solving eq (2) and (3) we get

$Q(1,-2)$

Slope of line $L_1$ is $m_1=1$ and of line $L_2$ is $m_1=-2$

$\tan \frac{\theta }{2}=\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

Let $\tan \frac{\theta }{2}=x$

$\frac{2x}{1-x^2}=\left|\frac{1+2}{1-2}\right|$

$\frac{2x}{1-x^2}=\left|\frac{-3}{1}\right|$

$\frac{2x}{1-x^2}=3$

$2x=3-3x^2$

$3x^2+2x-3=0$

$x=\frac{-2\pm \sqrt{2^2-4(-3)\left(3\right)}}{6}$

$x=\frac{-2\pm \sqrt{4+36}}{6}$

$x=\frac{-2\pm \sqrt{40}}{6}$

$x=\frac{-1\pm \sqrt{10}}{3}$

Now let $L_4$ be the angle bisector

$\Rightarrow m_4=\tan (\frac{\theta }{2}+{45}^0)=\frac{\frac{-1+\sqrt{10}}{3}+1}{1-\frac{\sqrt{10}-1}{3}}=\frac{2+\sqrt{10}}{4-\sqrt{10}}$

$R=(\frac{-2(4-\sqrt{10})}{2+\sqrt{10}},-2)$

$PR=\frac{-2(4-\sqrt{10})}{2+\sqrt{10}}+2=\frac{-4+4\sqrt{10}}{2+\sqrt{10}}$

$RQ=1+\frac{2(4-\sqrt{10})}{2+\sqrt{10}}=\frac{10-\sqrt{10}}{2+\sqrt{10}}$\

$\frac{PR}{RQ}=\frac{-4+4\sqrt{10}}{10-\sqrt{10}}$

$\frac{PR}{RQ}=\frac{2\sqrt{2}}{\sqrt{5}}$

Hence Statement I is correct

Statement -2

In any triangle, bisector of an angle divides the triangle into two similar triangle which is wrong

In $\Delta ABC$

if AD bisects $\angle A$

then,$\angle \frac{A}{2}=\angle \frac{A}{2}$

but we can't conclude about other two angles and sides