Question

Statement 1: The point $A(1,0,7)$ is the mirror image of the point $B(1,6,3)$ in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$
Statement-2: The line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ bisects the line segment joining $A(1,0,7)$ and $B(1,6,3)$

• A. Statement 1 is true, Statement 2 is false
• B. Statement 1 is false, Statement 2 is false
• C. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1
• D. Statement 1 is true, Statement 2 is true, Statement-2 is not a correct explanation for Statement 1

Answer 1

Answer: (C)

Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1

Consider the given Cartesian equation of line is $\frac{x}{1}=\frac{y-1}{2}=\frac{z-3}{3}......\left(1\right)$

Let the given point is $A(1,6,3)$

To find the image of $A(1,6,3)$ in the line draw a line $AB$ perpendicular to the line.

Let $C$ be the image of point $A$ and $B$.

Let $a,b,c$ be the direction cosines of $AC$is perpendicular to the line

We know that, condition of perpendicularity.

$a\times 1+b\times 2+c\times 3=0......\left(3\right)$

$a+2b+3c=0$

Now let, $\frac{x-1}{a}=\frac{y-6}{b}=\frac{z-3}{c}=k\left(say\right)......\left(3\right)$

Any point on the line (2) is $(ak+1,bk+6,ck+3)$

Let the point be $B$.

But $B$ also lies on the line $\left(1\right)$

$\therefore \frac{ak+1}{1}=\frac{bk+6-1}{2}=\frac{ck+3-2}{3}$

$\Rightarrow \frac{ak+1}{1}=\frac{bk+5}{2}=\frac{ck+1}{3}$

$\Rightarrow \frac{1(ak+1)+2(bk+5)+3(ck+1)}{1\times 1+2\times 2+3\times 3}$

$\Rightarrow \frac{14+(a+2b+3c)k}{14}=1$

$\Rightarrow (a+2b+3c)k=0$

$\Rightarrow ak=0,bk=-3,ck=2$

Then,

$B=(0+1,-3+6,2+3)$

$B=(1,3,5)$

Since, C is the midpoint of AB

$\frac{1+x^{\prime }}{2}=1and\frac{6+y^{\prime }}{2}=3and\frac{3+z^{\prime }}{2}=5$

$\Rightarrow x^{\prime }=1,y^{\prime }=0and{z}^{\prime }=7$

Hence $(1,0,7)$ is the image of A in line.