The locus of a point $P$ which divides the line joining $(1, 0)$ and $(2\cos \theta ,2\sin \theta )$ internally in the ratio $2 : 3$ for all $\theta $, is a
The locus of a point $P$ which divides the line joining $(1, 0)$ and $(2\cos \theta ,2\sin \theta )$ internally in the ratio $2 : 3$ for all $\theta $, is a
- [IIT 1986]
- A
Straight line
- B
Circle
- C
Pair of straight lines
- D
Parabola
Similar Questions
A pair of straight lines drawn through the origin form with the line $2x + 3y = 6$ an isosceles right angled triangle, then the lines and the area of the triangle thus formed is
A pair of straight lines drawn through the origin form with the line $2x + 3y = 6$ an isosceles right angled triangle, then the lines and the area of the triangle thus formed is
In a $\triangle A B C$, points $X$ and $Y$ are on $A B$ and $A C$, respectively, such that $X Y$ is parallel to $B C$. Which of the two following equalities always hold? (Here $[P Q R]$ denotes the area of $\triangle P Q R)$.
$I$. $[B C X]=[B C Y]$
$II$. $[A C X] \cdot[A B Y]=[A X Y] \cdot[A B C]$
In a $\triangle A B C$, points $X$ and $Y$ are on $A B$ and $A C$, respectively, such that $X Y$ is parallel to $B C$. Which of the two following equalities always hold? (Here $[P Q R]$ denotes the area of $\triangle P Q R)$.
$I$. $[B C X]=[B C Y]$
$II$. $[A C X] \cdot[A B Y]=[A X Y] \cdot[A B C]$
- [KVPY 2015]
Without using the Pythagoras theorem, show that the points $(4,4),(3,5)$ and $(-1,-1)$ are vertices of a right angled triangle.
Without using the Pythagoras theorem, show that the points $(4,4),(3,5)$ and $(-1,-1)$ are vertices of a right angled triangle.
The line $2x + 3y = 12$ meets the $x -$ axis at $A$ and the $y -$ axis at $B$ . The line through $(5, 5)$ perpendicular to $AB$ meets the $x -$ axis, $y -$ axis $\&$ the line $AB$ at $C, D, E$ respectively. If $O$ is the origin, then the area of the $OCEB$ is :
The line $2x + 3y = 12$ meets the $x -$ axis at $A$ and the $y -$ axis at $B$ . The line through $(5, 5)$ perpendicular to $AB$ meets the $x -$ axis, $y -$ axis $\&$ the line $AB$ at $C, D, E$ respectively. If $O$ is the origin, then the area of the $OCEB$ is :
An equilateral triangle has each of its sides of length $6\,\, cm$ . If $(x_1, y_1) ; (x_2, y_2) \,\, and \,\, (x_3, y_3)$ are its vertices then the value of the determinant,${{\left| {\,\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}\,} \right|}^2}$ is equal to :
An equilateral triangle has each of its sides of length $6\,\, cm$ . If $(x_1, y_1) ; (x_2, y_2) \,\, and \,\, (x_3, y_3)$ are its vertices then the value of the determinant,${{\left| {\,\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}\,} \right|}^2}$ is equal to :