Image of a Point in the Plane - Practice Questions & MCQ

The image of the point P(x₁, y₁, z₁) in the plane ax + by + cz + d = 0 is Q(x₃, y₃, z₃) then coordinates of point Q is given by

$\frac{\mathrm{x}_3-\mathrm{x}_1}{\mathrm{a}}=\frac{\mathrm{y}_3-\mathrm{y}_1}{\mathrm{~b}}=\frac{\mathrm{z}_3-\mathrm{z}_1}{\mathrm{c}}=-\frac{2\left(\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}\right)}{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}$

Proof:

Let point Q(x₃, y₃, z₃) is the image of the point P(x₁, y₁, z₁) in the plane π : ax + by + cz + d = 0

Let line PQ meets the plane ax+by+cz+d=0 at point R, direction ratio of normal to plane π are (a, b, c), since, PQ perpendicular to plane π.

So direction ratio of PQ are a, b, c 

$\Rightarrow$ equation of line PQ is
$
\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=r \quad \text { (say) }
$
any point on line $P Q$ may be taken as
Let
$
\begin{array}{r}
\left(a r+x_1, b r+y_1, c r+z_1\right) \\
Q \equiv\left(a r+x_1, b r+y_1, c r+z_1\right)
\end{array}
$
since, $R$ is the middle point of $P Q$
$
\begin{aligned}
& R \\
& \Rightarrow \quad \equiv\left(\frac{x_1+a r+x_1}{2}, \frac{y_1+b r+y_1}{2}, \frac{z_1+c r+z_1}{2}\right) \\
& R \\
& \equiv\left(x_1+\frac{a r}{2}, y_1+\frac{b r}{2}, z_1+\frac{c r}{2}\right)
\end{aligned}
$

Since, point R lies in the plane π, we get

$\begin{array}{cc}\Rightarrow & R \equiv\left(x_1+\frac{a r}{2}, y_1+\frac{b r}{2}, z_1+\frac{c r}{2}\right) \\ \Rightarrow & a\left(x_1+\frac{a r}{2}\right)+b\left(y_1+\frac{b r}{2}\right)+c\left(z_1+\frac{c r}{2}\right)+d=0 \\ \Rightarrow & \left(a^2+b^2+c^2\right) \frac{r}{2}=-\left(a x_1+b y_1+c z_1+d\right) \\ \Rightarrow & r=\frac{-2\left(a x_1+b y_1+c z_1+d\right)}{a^2+b^2+c^2}\end{array}$

In a similar method, we can also find the coordinates of point R.