Locus of points which are at equal distance from 3x + 4y - 11 = 0 and 12x + 5y + 2 = 0 and which is

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9.Straight Line

hard

Locus of the points which are at equal distance from $3x + 4y - 11 = 0$ and $12x + 5y + 2 = 0$ and which is near the origin is

$21x - 77y + 153 = 0$

$99x + 77y - 133 = 0$

$7x - 11y = 19$

None of these

Solution

(b) Let point be $({x_1},{y_1}),$then according to the condition $\frac{{3{x_1} + 4{y_1} – 11}}{5} = – \left( {\frac{{12{x_1} + 5{y_1} + 2}}{{13}}} \right)$

Since the given lines are on opposite sides with respect to origin, hence the required locus is $99x + 77y – 133 = 0$ .

Standard 11

Mathematics

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 :

normal

The equation of the lines on which the perpendiculars from the origin make ${30^o}$ angle with $x$-axis and which form a triangle of area $\frac{{50}}{{\sqrt 3 }}$ with axes, are

medium

Let the points of intersections of the lines $x-y+1=0$, $x-2 y+3=0$ and $2 x-5 y+11=0$ are the mid points of the sides of a triangle $A B C$. Then the area of the triangle $\mathrm{ABC}$ is …. .

hard

(JEE MAIN-2021)

Let $A B C D$ be a tetrahedron such that the edges $AB , AC$ and $AD$ are mutually perpendicular. Let the areas of the triangles $ABC , ACD$ and $ADB$ be $5,6$ and $7$ square units respectively. Then the area (in square units) of the $\triangle BCD$ is equal to :

hard

(JEE MAIN-2025)

One vertex of the equilateral triangle with centroid at the origin and one side as $x + y – 2 = 0$ is

hard

Locus of the points which are at equal distance from $3x + 4y - 11 = 0$ and $12x + 5y + 2 = 0$ and which is near the origin is

Solution

Similar Questions

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 equation of the lines on which the perpendiculars from the origin make ${30^o}$ angle with $x$-axis and which form a triangle of area $\frac{{50}}{{\sqrt 3 }}$ with axes, are

Let the points of intersections of the lines $x-y+1=0$, $x-2 y+3=0$ and $2 x-5 y+11=0$ are the mid points of the sides of a triangle $A B C$. Then the area of the triangle $\mathrm{ABC}$ is …. .

Let $A B C D$ be a tetrahedron such that the edges $AB , AC$ and $AD$ are mutually perpendicular. Let the areas of the triangles $ABC , ACD$ and $ADB$ be $5,6$ and $7$ square units respectively. Then the area (in square units) of the $\triangle BCD$ is equal to :

One vertex of the equilateral triangle with centroid at the origin and one side as $x + y – 2 = 0$ is

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