Image of a Point in the given Line - Practice Questions & MCQ

Since P (foot of perpendicular) is the midpoint of M and N (image of a point M in the line), we can get N if P is found out. 

Cartesian Form

Let $\mathbf{M}(\boldsymbol{\alpha}, \boldsymbol{\beta}, \mathrm{\gamma})$ be the point and $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$ be the equation of line $\mathbf{L}$.
Let $\mathbf{P}$ be the foot of the perpendicular from $\mathbf{M}$ to line $\mathbf{L}$ and let $\mathbf{N}$ be the image of the point in the given line, where $\mathbf{M P}=\mathbf{P N}$
Let the coordinates of $P$ be
$
\left(x_0+a \lambda, y_0+b \lambda, z_0+c \lambda\right)
$

Then, direction ratios of $P L$ are
$
\left(x_0+a \lambda-\alpha, y_0+b \lambda-\beta, z_0+c \lambda-\gamma\right)
$
since, MP is perpendicular to the given line, whose direction ratios are $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$.
$
\begin{array}{ll}
\therefore & a \cdot\left(x_0+a \lambda-\alpha\right)+b \cdot\left(y_0+b \lambda-\beta\right)+c \cdot\left(z_0+c \lambda-\gamma\right)=0 \\
\Rightarrow & \lambda=\frac{a\left(\alpha-x_0\right)+b\left(\beta-y_0\right)+c\left(\gamma-z_0\right)}{a^2+b^2+c^2}
\end{array}
$

Substituting the value of λ we get coordinates of point P (foot of perpendicular)

As N (α’, β’, γ’) is image of point M(α, β, γ)

∴ mid-point of MN is point P
$
\begin{array}{ll}
\therefore & \frac{\alpha+\alpha^{\prime}}{2}=x_0+a \lambda, \quad \frac{\beta+\beta^{\prime}}{2}=y_0+b \lambda, \quad \frac{\gamma+\gamma^{\prime}}{2}=z_0+\alpha \lambda \\
\therefore & \alpha^{\prime}=2\left(x_0+a \lambda\right)-\alpha, \beta^{\prime}=2\left(y_0+b \lambda\right)-\beta, \gamma^{\prime}=2\left(z_0+c \lambda\right)-\gamma
\end{array}
$