Animated Solution for Mathematics - Three Dimensional Geometry: The image of the point $(-1,3,4)$ in the plane $x-2y=0$ is

Imagine you are standing in a room, and before you hangs a perfectly flat, infinite mirror. In our problem, this mirror is a plane defined by the equation $x-2y=0$.

You are holding a point $P(-1,3,4)$ in your hand, and you want to see where its reflection, $P^{\prime }$, would appear in this mirror. This is a fundamental exploration of symmetry in three-dimensional space.

To find the image $P^{\prime }(x^{\prime },y^{\prime },z^{\prime })$, we must understand the path of light. The line connecting the object $P$ and its image $P^{\prime }$ must be perpendicular to the plane.

This means the line is parallel to the plane's normal vector, $\vec{n}=(1,-2,0)$. The plane acts as the perpendicular bisector of the segment $P{P}^{\prime }$.

Our task is to find the point $P^{\prime }$ such that the midpoint of $P{P}^{\prime }$ lies on the plane and the vector $\overrightarrow{P{P}^{\prime }}$ is parallel to $\vec{n}$.

We use the elegant standard formula for the image of a point $(x_1,y_1,z_1)$ in a plane $ax+by+cz+d=0$:

This formula is a masterpiece of vector geometry. The term on the right is a constant ratio, which encapsulates the distance from the point to the plane.

The $-2$ is the key—it ensures we jump from the point, through the plane, and land exactly the same distance on the other side.

First, we identify our parameters: $x_1=-1$, $y_1=3$, $z_1=4$, and the plane coefficients $a=1$, $b=-2$, $c=0$, $d=0$.

Now, let's calculate the constant ratio $k$:

Simplifying the numerator: $1(-1)-2(3)=-1-6=-7$. The denominator is $1^2+(-2{)}^2+0^2=1+4=5$.

Thus, $k=-2\left(\frac{-7}{5}\right)=\frac{14}{5}$.

With $k$ in hand, we find the coordinates:

For $x^{\prime }$:

For $y^{\prime }$:

For $z^{\prime }$:

Since the denominator is zero, the numerator must be zero, so $z^{\prime }-4=0$, which gives $z^{\prime }=4$.

Our calculated image point is $P^{\prime }(\frac{9}{5},-\frac{13}{5},4)$.

In the high-stakes environment of the JEE, this is the moment where confidence in your process is your greatest asset. Do not doubt your math; trust the geometry.

The correct answer is 'None of these'. You have successfully navigated the mirror, and the physics of the reflection remains perfectly consistent.