A square of side a lies above x -axis and has one vertex at origin. side passing through origin make

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9.Straight Line

hard

A square of side a lies above the $x$ -axis and has one vertex at the origin. The side passing through the origin makes an angle $\alpha ,(0 < \alpha < \frac{\pi }{4})$ with the positive direction of $x$-axis. The equation of its diagonal not passing through the origin is

$y(\cos \alpha - \sin \alpha ) - x(\sin \alpha - \cos \alpha ) = a$

$y(\cos \alpha + \sin \alpha ) - x(\sin \alpha - \cos \alpha ) = a$

$y(\cos \alpha + \sin \alpha ) + x(\sin \alpha + \cos \alpha ) = a$

$y(\cos \alpha + \sin \alpha ) + x(\sin \alpha - \cos \alpha ) = a$

(AIEEE-2003)

Solution

(b) Co-ordinates of $A = (a\cos \alpha ,\,a\sin \alpha )$
Equation of $OB$, $y = \tan \left( {\frac{\pi }{4} + \alpha } \right)\,x$
$\because$ $CA\,\,{ \bot ^r}\,\,{\rm{to}}\,\,OB$
$\therefore$ Slope of $CA = – \cot \left( {\frac{\pi }{4} + \alpha } \right)$
Equation of CA,$y – a\sin \alpha = – \cot \left( {\frac{\pi }{4} + \alpha } \right)\,(x – a\cos \alpha )$
==> $y(\sin \alpha + \cos \alpha ) + x(\cos \alpha – \sin \alpha ) = a$
$ \Rightarrow $ $y\,(\cos \alpha + \sin \alpha ) – x\,(\sin \alpha – \cos \alpha ) = a$.

Standard 11

Mathematics

A rod of length eight units moves such that its ends $A$ and $B$ always lie on the lines $x-y+2=0$ and $y+2=0$, respectively. If the locus of the point $P$, that divides the $\operatorname{rod} AB$ internally in the ratio $2: 1$ is $9\left(x^2+\alpha y^2+\beta x y+\gamma x+28 y\right)-76=0$, then $\alpha-\beta-\gamma$ is equal to :

hard

(JEE MAIN-2025)

Consider the lines $L_1$ and $L_2$ defined by

$L _1: x \sqrt{2}+ y -1=0$ 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 $L_1$ and the distance of $P$ from $L_2$ is $\lambda^2$. The line $y=2 x+1$ 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 $R ^{\prime}$ and $S ^{\prime}$. Let $D$ be the square of the distance between $R ^{\prime}$ and $S ^{\prime}$.

($1$) The value of $\lambda^2$ is

($2$) The value of $D$ is

normal

(IIT-2021)

Let $A B C$ be a triangle formed by the lines $7 x-6 y+3=0, x+2 y-31=0$ and $9 x-2 y-19=0$, Let the point $( h , k )$ be the image of the centroid of $\Delta A B C$ in the line $3 x+6 y-53=0$. Then $h^2+k^2+h k$ is equal to

hard

(JEE MAIN-2025)

Let $PS$ be the median of the triangle with vertices $P(2,2) , Q(6,-1) $ and $R(7,3) $. The equation of the line passing through $(1,-1) $ and parallel to $PS $ is :

medium

(IIT-2000)

Two sides of a parallelogram are along the lines $4 x+5 y=0$ and $7 x+2 y=0$. If the equation of one of the diagonals of the parallelogram is $11 \mathrm{x}+7 \mathrm{y}=9$, then other diagonal passes through the point:

hard

(JEE MAIN-2021)

A square of side a lies above the $x$ -axis and has one vertex at the origin. The side passing through the origin makes an angle $\alpha ,(0 < \alpha < \frac{\pi }{4})$ with the positive direction of $x$-axis. The equation of its diagonal not passing through the origin is

Solution

Similar Questions

Let $A B C$ be a triangle formed by the lines $7 x-6 y+3=0, x+2 y-31=0$ and $9 x-2 y-19=0$, Let the point $( h , k )$ be the image of the centroid of $\Delta A B C$ in the line $3 x+6 y-53=0$. Then $h^2+k^2+h k$ is equal to

Let $PS$ be the median of the triangle with vertices $P(2,2) , Q(6,-1) $ and $R(7,3) $. The equation of the line passing through $(1,-1) $ and parallel to $PS $ is :

Two sides of a parallelogram are along the lines $4 x+5 y=0$ and $7 x+2 y=0$. If the equation of one of the diagonals of the parallelogram is $11 \mathrm{x}+7 \mathrm{y}=9$, then other diagonal passes through the point:

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