The lines L_1: y-x=0 and L_2: 2x+y=0 intersect the line L_3: y+2=0 at P and Q respectively…

Question

The lines $L_1: y-x=0$ and $L_2: 2x+y=0$ 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 $PR:RQ$ equals $2\sqrt2:\sqrt5$

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

Accepted Answer

${L_1}:y - x = 0$

${L_2}:2x + y = 0$

${L_3}:y + 2 = 0$

On solving the equation of line ${L_1}$ and ${L_2}$ we get their point of

intersection $(0, 0)$ i.e., origin $O.$

On solving the equation of line ${L_1}$ and ${L_3},$

we get $P=(-2, -2).$

Similarly, we get $$Q = \left( { - 1, - 2} \right)$$

We know that bisector of an angle of a triangle, divide the opposite side the triangle in the ratio of the sides including the angle [ Angle Bisector Theorem of a Triangle ]

$$\therefore$$ $${{PR} \over {RQ}} = {{OP} \over {OQ}} = {{\sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 2} \right)}^2}} } \over {\sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 2} \right)}^2}} }}$$

$$ = {{2\sqrt 2 } \over {\sqrt 5 }}$$

Solutions · 1