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 $$2\sqrt 2 :\sqrt 5 $$
Statement-2: In any triangle, bisector of an angle divide the triangle into two similar triangles.
A
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
B
Statement-1 is true, Statement-2 is false.
C
Statement-1 is false, Statement-2 is true.
D
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Explanation
$${L_1}:y - x = 0$$
$${L_2}:2x + y = 0$$
$${L_3}:y + 2 = 0$$
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 ]
The line $$L$$ given by $${x \over 5} + {y \over b} = 1$$ passes through the point $$\left( {13,32} \right)$$. The line K is parrallel to $$L$$ and has the equation $${x \over c} + {y \over 3} = 1.$$ Then the distance between $$L$$ and $$K$$ is
Three distinct points A, B and C are given in the 2 -dimensional coordinates plane such that the ratio of the distance of any one of them from the point $$(1, 0)$$ to the distance from the point $$(-1, 0)$$ is equal to $${1 \over 3}$$. Then the circumcentre of the triangle ABC is at the point:
$$\therefore$$ A lies on the circle given by eq. $$(1).$$ As $$B$$ and $$C$$
also follow the same condition, - they must lie on the same circle.
$$\therefore$$ Center of circumcircle of $$\Delta ABC$$
$$=$$ Center of circle given by $$\left( 1 \right) = \left( {{5 \over 4},0} \right)$$
4
AIEEE 2009
MCQ (Single Correct Answer)
The lines $$p\left( {{p^2} + 1} \right)x - y + q = 0$$ and $$\left( {{p^2} + 1} \right){}^2x + \left( {{p^2} + 1} \right)y + 2q$$ $$=0$$ are perpendicular to a common line for :
A
exactly one values of $$p$$
B
exactly two values of $$p$$
C
more than two values of $$p$$
D
no value of $$p$$
Explanation
If the lines $$p\left( {{p^2} + 1} \right)x - y + q = 0$$