Locus of the image of point (2,3) in the lin | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Let $M$ be mid-point of $B B^{\prime}$ and $A M$ is $\perp$ bisector of $B B^{\prime}$

(where $A$ is the point of intersection of the given lines)

Vedclass · Accepted Answer

circle of radius  $\;\sqrt 2 $

Vedclass · Answer

Let $M$ be mid-point of $B B^{\prime}$ and $A M$ is $\perp$ bisector of $B B^{\prime}$

(where $A$ is the point of intersection of the given lines)

$(x-2)(x-1)+(y-2)(y-3)=0$

$\Rightarrow\left(\frac{h+2}{2}-2\right)\left(\frac{h+2}{2}-1\right)+\left(\frac{k+3}{2}-2\right)\left(\frac{k+3}{2}-3\right)=0$

$\Rightarrow(h-2)(h)+(k-1)(k-3)=0$

$\Rightarrow \quad x^{2}-2 x+y^{2}-4 y+3=0$

$\Rightarrow \quad(x-1)^{2}+(y-2)^{2}=2$

Locus of the image of point $ (2,3)$ in the line $\left( {2x - 3y + 4} \right) + k\left( {x - 2y + 3} \right) = 0,k \in R$ is a:

Similar Questions

If the straight lines $x + 3y = 4,\,\,3x + y = 4$ and $x +y = 0$ form a triangle, then the triangle is

The area enclosed within the curve $|x| + |y| = 1$ is

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

If in a parallelogram $ABDC$, the coordinates of $A, B$ and $C$ are respectively $(1, 2), (3, 4)$ and $(2, 5)$, then the equation of the diagonal $AD$ 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: