Question

Locus of the image of the point (2,3) in the line $2x-3y+4+k\left( x-2y+3 \right)=0$ , $k\in R$ , is a:
(a) Straight line parallel to x-axis
(b) Straight line parallel to y-axis
(c) Circle of radius $\sqrt{2}$
(d) Circle of radius $\sqrt{3}$

Answer 1

Hint: Let the reflection of the point (2,3) be (h,g). So, the points (2,3) and (h,g) must be at equal distance from the meet point of the line represented by the equation $2x-3y+4+k\left( x-2y+3 \right)=0$ ; which can be found by finding the meet point of the lines $2x-3y+4=0$ and $x-2y+3=0$

Complete step-by-step answer:

Let us start the solution to the above question by letting the reflection of the point (2,3) be (h,g).

So, the points (2,3) and (h,g) must be at equal distance from the meet point of the line represented by the equation $2x-3y+4+k\left( x-2y+3 \right)=0$ ; which can be found by finding the meet point of the lines $2x-3y+4=0$ and $x-2y+3=0$ . So, let us find the point of intersection:

$2x-3y+4=0.............(i)$

$x-2y+3=0..........(ii)$

If we subtract 2 times equation (ii) from equation (i), we get

$2x-3y+4-2\left( x-2y+3 \right)=0$

$\Rightarrow 2x-3y+4-2x+4y-6=0$

$\Rightarrow y-2=0$

$\Rightarrow y=2$

If we put this y in equation (ii), we get

$x-2y+3=0$

$\Rightarrow x-4+3=0$

$\Rightarrow x=1$

So, the intersection of the lines represented by the equation $2x-3y+4+k\left( x-2y+3 \right)=0$ is (1,2).

Now, let us use the distance formula and equate the distance of the points (2,3) and (h,g) from (1,2). According to the distance formula, the distance between two points $\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ .

$\sqrt{{{\left( h-1 \right)}^{2}}+{{\left( g-2 \right)}^{2}}}=\sqrt{{{\left( 2-1 \right)}^{2}}+{{\left( 3-2 \right)}^{2}}}$

If we square both sides of the equation, we get

${{\left( h-1 \right)}^{2}}+{{\left( g-2 \right)}^{2}}={{\left( 2-1 \right)}^{2}}+{{\left( 3-2 \right)}^{2}}$

$\Rightarrow {{\left( h-1 \right)}^{2}}+{{\left( g-2 \right)}^{2}}={{1}^{2}}+{{1}^{2}}$

$\Rightarrow {{\left( h-1 \right)}^{2}}+{{\left( g-2 \right)}^{2}}=2$

Now, if we substitute h=x and g=y, we get

${{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}=2$

So, the above equation represents the equation of a circle with radius $\sqrt{2}$ , which we can say by comparing the equation with the equation of circle of radius r represented by ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$ .

The locus can be represented in the form of a diagram as:

So, the correct answer is “Option C”.

Note: If you want you can solve the above question using the formula of reflection of point about a line according to which we know that the image (r,s) of a point (p,q) to a general line ax+by+c=0 is given by: $\dfrac{r-p}{a}=\dfrac{s-q}{b}=\dfrac{-2\left( ap+bq+c \right)}{{{a}^{2}}+{{b}^{2}}}$ . Also, the centre of the circle represented by ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$ is a (a,b).