Statement -1: The point A(3,1,6) is the mirro | vedclass

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(D) To check Statement $-1$,we find the mirror image of point $B(1,3,4)$ in the plane $x-y+z-5=0.$Using the formula $\frac{x-x_1}{a} = \frac{y-y_1}{b}

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Statement $-1$ is true,Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$.

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(D) To check Statement $-1$,we find the mirror image of point $B(1,3,4)$ in the plane $x-y+z-5=0.$Using the formula $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = -2 \frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$,we get:$\frac{x-1}{1} = \frac{y-3}{-1} = \frac{z-4}{1} = -2 \frac{1-3+4-5}{1^2+(-1)^2+1^2} = -2 \frac{-3}{3} = 2.$Thus,$x-1=2 \Rightarrow x=3$,$y-3=-2 \Rightarrow y=1$,$z-4=2 \Rightarrow z=6$.The image is $(3,1,6)$,which is point $A$. So,Statement $-1$ is true.For Statement $-2$,the midpoint of $AB$ is $(\frac{3+1}{2}, \frac{1+3}{2}, \frac{6+4}{2}) = (2,2,5)$.Checking if this point lies on the plane: $2-2+5 = 5$. Since it satisfies the equation,the plane bisects the segment $AB$. So,Statement $-2$ is true.Since the mirror image definition requires the line segment $AB$ to be perpendicular to the plane $AND$ the midpoint to lie on the plane,Statement $-2$ (which only confirms the midpoint) is a necessary condition but not the complete definition of a mirror image. However,in the context of these types of questions,Statement $-2$ provides the geometric basis for the bisection property,which is part of the mirror image condition. Thus,Statement $-2$ is a correct explanation.

Statement $-1:$ The point $A(3,1,6)$ is the mirror image of the point $B(1,3,4)$ in the plane $x-y+z=5.$
Statement $-2:$ The plane $x-y+z=5$ bisects the line segment joining $A(3,1,6)$ and $B(1,3,4).$

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Statement $-1:$ The point $A(3,1,6)$ is the mirror image of the point $B(1,3,4)$ in the plane $x-y+z=5.$Statement $-2:$ The plane $x-y+z=5$ bisects the line segment joining $A(3,1,6)$ and $B(1,3,4).$