Straight Lines and Pair of Straight Lines · Mathematics · JEE Advanced
Numerical
$${L_1}:x\sqrt 2 + y - 1 = 0$$ and $${L_2}:x\sqrt 2 - y + 1 = 0$$
For a fixed constant $$\lambda$$, let C be the locus of a point P such that the product of the distance of P from L1 and the distance of P from L2 is $$\lambda$$2. The line y = 2x + 1 meets C at two points R and S, where the distance between R and S is $$\sqrt {270} $$. Let the perpendicular bisector of RS meet C at two distinct points R' and S'. Let D be the square of the distance between R' and S'.
The value of $$\lambda$$2 is __________.
$${L_1}:x\sqrt 2 + y - 1 = 0$$ and $${L_2}:x\sqrt 2 - y + 1 = 0$$
For a fixed constant $$\lambda$$, let C be the locus of a point P such that the product of the distance of P from L1 and the distance of P from L2 is $$\lambda$$2. The line y = 2x + 1 meets C at two points R and S, where the distance between R and S is $$\sqrt {270} $$. Let the perpendicular bisector of RS meet C at two distinct points R' and S'. Let D be the square of the distance between R' and S'.
The value of D is __________.
MCQ (Single Correct Answer)
Consider three points $$P = ( - \sin (\beta - \alpha ), - cos\beta ),Q = (cos(\beta - \alpha ),\sin \beta )$$ and $$R = (\cos (\beta - \alpha + \theta ),\sin (\beta - \theta ))$$ where $$0 < \alpha ,\beta ,\theta < {\pi \over 4}$$. Then :
Consider the lines given by:
$${L_1}:x + 3y - 5 = 0$$
$${L_2}:3x - ky - 1 = 0$$
$${L_3}:5x + 2y - 12 = 0$$
Match the Statement/Expressions in Column I with the Statements/Expressions in Column II.
Column I | Column II | ||
---|---|---|---|
(A) | L$$_1$$, L$$_2$$, L$$_3$$ are concurrent, if | (P) | $$K = - 9$$ |
(B) | One of L$$_1$$, L$$_2$$, L$$_3$$ is parallel to atleast one of the other two, if | (Q) | $$K = - {6 \over 5}$$ |
(C) | L$$_1$$, L$$_2$$, L$$_3$$ form a triangle, if | (R) | $$K = {5 \over 6}$$ |
(D) | L$$_1$$, L$$_2$$, L$$_3$$ do not form a triangle, if | (S) | $$K = 5$$ |
Let a and b be non-zero real numbers. Then, the equation
$$(a{x^2} + b{y^2} + c)({x^2} - 5xy + 6{y^2}) = 0$$ represents :
Statement-1: The ratio $$PR$$ : $$RQ$$ equals $$2\sqrt 2 :\sqrt 5 $$. because
Statement-2: In any triangle, bisector of an angle divides the triangle into two similar triangles.
Then the equation of the bisector of the angle $$PQR$$ is
$$3a{x^2} + 5xy + \left( {{a^2} - 2} \right){y^2} = 0$$ are perpendicular to each other for
Reflection about the line $$y=x$$.
Translation through a distance 2 units along the positive direction of x-axis.
Rotation through an angle $$p/4$$ about the origin in the counter clockwise direction.
Then the final position of the point is given by the coordinates.
MCQ (More than One Correct Answer)
Subjective
the equation $$\left| {\matrix{ {ax - by - c} & {bx + ay} & {cx + a} \cr {bx + ay} & { - ax + by - c} & {cy + b} \cr {cx + a} & {cy + b} & { - ax - by + c} \cr } } \right| = 0$$
represents a straight line.
[area $$\left( {\Delta {P_1},{P_2},{P_3}} \right)$$]/[area $$\left( {{P_2},{P_3},{P_4}} \right)$$]
$$\left[ {a{t_1}{t_2},\,\,a\left( {{t_1} + {t_2}} \right)} \right],\,\,\left[ {a{t_2}{t_3},a\left( {{t_2} + {t_3}} \right)} \right],\,\,\left[ {a{t_3}{t_1},\,a\left( {{t_3} + {t_1}} \right)} \right]$$. Find the orthocentre of the triangle.
(b) Find the equation of the line which bisects the obtuse angle between the lines $$x - 2y + 4 = 0$$ and $$4x - 3y + 2 = 0$$.