Foot of Perpendicular and Image

In geometry, the concept of the "foot of perpendicular" refers to the point at which a perpendicular line drawn from a point meets another line, plane, or surface. The foot of the perpendicular is widely used in determining distances, angles, and projections, making it an essential tool for solving numerous geometric problems.

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Foot of Perpendicular

The foot of the perpendicular is defined as the point at which a perpendicular line from a given point intersects another geometric object, such as a line, plane, or surface. If P is a point not on the line L, and a perpendicular is dropped from P to L, then the point F where the perpendicular meets L is called the foot of the perpendicular from P to L.

Foot of perpendicular of $P(x_1,y_1)$ on the line $AB$ : $ax+by+c=0$ is $M(x_2,y_2)$. then
$\frac{{\mathrm{x}}_2-{\mathrm{x}}_1}{\mathbf{a}}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{\mathrm{b}}=-\frac{({\mathrm{ax}}_1+{\mathrm{by}}_1+\mathrm{c})}{({\mathrm{a}}^2+{\mathrm{b}}^2)}$

Proof:
Let the coordinate of the foot of the perpendicular be $M(x_2,y_2)$. Then, point $M$ lies on the line $AB$.
$\Rightarrow {\mathrm{ax}}_2+{\mathrm{by}}_2+\mathrm{c}=0$
and, $\because \mathrm{PM}\perp \mathrm{AB}$
then, (slope of PM$)($ slope of AB$)=-1$
$\begin{array}{ll}\Rightarrow & \left(\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}\right)(-\frac{\mathrm{a}}{\mathrm{b}})=-1\\ \text{ or }& \frac{{\mathrm{x}}_2-{\mathrm{x}}_1}{\mathrm{a}}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{\mathrm{b}}\end{array}$

Using ratio and proportion
$\begin{array}{rl}& \frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=\frac{a(x_2-x_1)+b(y_2-y_1)}{a^2+b^2}\\ & \frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=-\frac{(a{x}_1+b{y}_1+c)}{a^2+b^2}\end{array}$
(Using (i))

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Solved Examples Based on Foot of Perpendicular:

Example 1: The coordinates of the foot of the perpendicular from the point $\mathbf{P}(\mathbf{2},\mathbf{3})$ to the line $3x-4y-16=0$
1) $(5,-1)$
2) $(3,4)$
3) $(1,2)$
4) $(2,3)$

Solution
$\frac{{\mathrm{x}}_2-{\mathrm{x}}_1}{\mathrm{a}}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{\mathrm{b}}=-\frac{({\mathrm{ax}}_1+{\mathrm{by}}_1+\mathrm{c})}{({\mathrm{a}}^2+{\mathrm{b}}^2)}$

Applying this formula
$\begin{array}{rl}& \frac{x_2-2}{3}=\frac{y_2-3}{-4}=-\frac{(3(2)-4(3)-19)}{3^2+4^2}\\ & \frac{x_2-2}{3}=\frac{y_2-3}{-4}=1\\ & \frac{x_2-2}{3}=1\text{ and }\frac{y_2-3}{-4}=1\\ & x_2=5\text{ and }{y}_2=-1\end{array}$

Hence, the answer is the option 1.

Example 2: The coordinates of the foot of perpendicular from $(\mathrm{a},0)$ on the line $y=mx+\frac{a}{m}$ are
1) $(0,\frac{-a}{m})$
2) $(\frac{a}{m},0)$
3) $(0,\frac{a}{m})$
4) None

Solution
The line can be re-written as $mx-y+\frac{a}{m}=0$
Using the formula
$\begin{array}{rl}& \frac{{\mathbf{x}}_2-{\mathbf{x}}_{\mathbf{1}}}{\mathbf{a}}=\frac{{\mathbf{y}}_{\mathbf{2}}-{\mathbf{y}}_{\mathbf{1}}}{\mathbf{b}}=-\frac{({\mathbf{a}\mathbf{x}}_{\mathbf{1}}+{\mathbf{b}\mathbf{y}}_{\mathbf{1}}+\mathbf{c})}{({\mathbf{a}}^2+{\mathbf{b}}^2)}\\ & \frac{x_2-a}{m}=\frac{y_2-0}{-1}=-\frac{m(a)-0+\frac{a}{m}}{m^2+1^2}\\ & \frac{x_2-a}{m}=\frac{y_2}{-1}=-\frac{a}{m}\\ & x_2=0\text{ and }{y}_2=\frac{a}{m}\end{array}$

Hence, the answer is the option 3.

Example 3: The foot of the perpendicular on the line $4\mathrm{x}+\mathrm{y}=\mathrm{k}$ drawn from the origin is C . If the line cuts the x -axis and Y -axis at A and B respectively then BC : CA is
1) $1:4$
2) $4:1$
3) $1:16$
4) $16:1$

Solution
$\tan ({180}^{\circ }-\theta )=$ slope of $\mathrm{AB}=-4$
$\therefore \tan \theta =4$
$\therefore \frac{\mathrm{OC}}{\mathrm{AC}}=\tan \theta ,\frac{\mathrm{OC}}{\mathrm{BC}}=\cot \theta$
$\Rightarrow \frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\tan \theta }{\cot \theta }={\tan }^2\theta =16$

Example 4: The foot of the perpendicular on the line $x+3y=\lambda$ drawn from the origin is C. If the line cuts the $x$-axis a $y$-axis at A and B respectively then $\mathrm{BC}:\mathrm{CA}$ is

1) $1:3$
2) $3:1$
3) $1:9$
4) $9:1$

Solution
$\tan ({180}^{\circ }-\theta )=$ slope of $\mathrm{AB}=-1/3$
$\therefore \tan \theta =1/3$
$\therefore \frac{\mathrm{OC}}{\mathrm{AC}}=\tan \theta ,\frac{\mathrm{OC}}{\mathrm{BC}}=\cot \theta$
$\Rightarrow \frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\tan \theta }{\cot \theta }={\tan }^2\theta =1/9$

Example 5: The middle point of the line segment joining $(3,-1)$ and $(1,1)$ is shifted by two units (in the sense of increasing $y$ ) perpendicular to the line segment. Find the co-ordinates of the point in the new position.
1) $(4,\sqrt{2})$
2) $(2+\sqrt{2},\sqrt{2})$
3) $(2+\sqrt{2},2)$
4) $(2-\sqrt{2},2)$

Solution
Let $P$ be the middle point of the line segment joining
$\mathrm{A}(3,-2)$ and $\mathrm{B}(1,1)$
Then
$\mathrm{P}=(\frac{3+1}{2},\frac{-1+1}{2})=(2,0)$

Let $P$ be shifted to $Q$ where $PQ=2$ and $y$ co-ordinate of $Q$ is greater than $P$
Now slope of $AB=-1$
$\therefore$ slope of $\mathrm{PQ}=1$

Co-ordinates of Q by distance formula $=(2\pm 2\cos \theta ,0\pm 2\sin \theta )$ where $\tan \theta =1$
$=(2\pm 2\cdot \frac{1}{\sqrt{2}},0\pm 2\cdot \frac{1}{\sqrt{2}})=(2\pm \sqrt{2},\pm \sqrt{2})$
as $y$ co-ordinates of $Q$ is greater than that of $P$.
Hence, $\mathrm{Q}=(2+\sqrt{2},\sqrt{2})$ is the required point.