Foot of Perpendicular and Image

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

Proof:
Let the coordinate of the foot of the perpendicular be $M\left(x_2, y_2\right)$. Then, point $M$ lies on the line $A B$.
$
\Rightarrow \quad \mathrm{ax}_2+\mathrm{by}_2+\mathrm{c}=0
$
and, $\quad \because \mathrm{PM} \perp \mathrm{AB}$
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)\left(-\frac{\mathrm{a}}{\mathrm{b}}\right)=-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{aligned}
& \frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=\frac{a\left(x_2-x_1\right)+b\left(y_2-y_1\right)}{a^2+b^2} \\
& \frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=-\frac{\left(a x_1+b y_1+c\right)}{a^2+b^2}
\end{aligned}
$
(Using (i))

Recommended Video Based on Foot of Perpendicular

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 $3 x-4 y-16=0$
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{\left(\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}\right)}{\left(\mathrm{a}^2+\mathrm{b}^2\right)}
$

Applying this formula
$
\begin{aligned}
& \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{aligned}
$

Hence, the answer is the option 1.

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

Solution
The line can be re-written as $m x-y+\frac{a}{m}=0$
Using the formula
$
\begin{aligned}
& \frac{\mathbf{x}_2-\mathbf{x}_{\mathbf{1}}}{\mathbf{a}}=\frac{\mathbf{y}_{\mathbf{2}}-\mathbf{y}_{\mathbf{1}}}{\mathbf{b}}=-\frac{\left(\mathbf{a x}_{\mathbf{1}}+\mathbf{b y}_{\mathbf{1}}+\mathbf{c}\right)}{\left(\mathbf{a}^2+\mathbf{b}^2\right)} \\
& \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{aligned}
$

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

Example 4: The foot of the perpendicular on the line $x+3 y=\lambda$ 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
$\tan \left(180^{\circ}-\theta\right)=$ slope of $\mathrm{AB}=-1 / 3$
$\therefore \quad \tan \theta=1 / 3$
$\therefore \quad \frac{\mathrm{OC}}{\mathrm{AC}}=\tan \theta, \frac{\mathrm{OC}}{\mathrm{BC}}=\cot \theta$
$\Rightarrow \quad \frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\tan \theta}{\cot \theta}=\tan ^2 \theta=1 / 9$

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}=\left(\frac{3+1}{2}, \frac{-1+1}{2}\right)=(2,0)
$

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

Co-ordinates of Q by distance formula $=(2 \pm 2 \cos \theta, 0 \pm 2 \sin \theta)$ where $\tan \theta=1$
$
=\left(2 \pm 2 \cdot \frac{1}{\sqrt{2}}, \quad 0 \pm 2 \cdot \frac{1}{\sqrt{2}}\right)=(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.