Find the image of the point (0, 2, 3) in the line
Find the image of the point (0, 2, 3) in the line

Find the image of the point in the line .
Given: Equation of line is .
To find: image of point
Formula Used: Equation of a line is
Vector form:
Cartesian form:
where is a point on the line and with being the direction ratios of the line.
If 2 lines of direction ratios and are perpendicular, then Mid-point of line segment joining and is

$\left(\frac{{x}_{1}–{x}_{2}}{2},\frac{{y}_{1}–{y}_{2}}{2},\frac{{z}_{1}–{z}_{2}}{2}\right)$
\left(\frac{x_{1}-x_{2}}{2}, \frac{y_{1}-y_{2}}{2}, \frac{z_{1}-z_{2}}{2}\right)

Explanation:
Let

$\frac{x+3}{5}=\frac{y–1}{2}=\frac{z+4}{3}=\lambda$
\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=\lambda

So the foot of the perpendicular is
The direction ratios of the perpendicular is

$\begin{array}{l}\left(5\lambda –3–0\right):\left(2\lambda +1–2\right):\left(3\lambda –4–3\right)\\ ⇒\left(5\lambda –3\right):\left(2\lambda –1\right):\left(3\lambda –7\right)\end{array}$
\begin{array}{l}
(5 \lambda-3-0):(2 \lambda+1-2):(3 \lambda-4-3) \\
\Rightarrow(5 \lambda-3):(2 \lambda-1):(3 \lambda-7)
\end{array}

Direction ratio of the line is