The coordinates of the foot of perpendiculars from the point $(2,3)$ on the line $y=3x+4$ is given by (A) $\left(\dfrac{37}{10},\dfrac{-1}{10}\right)$ (B) $\left(\dfrac{-1}{10},\dfrac{37}{10}\right)$ (C) $\left(\dfrac{10}{37},-10\right)$ (D) $\left(\dfrac{2}{3},-\dfrac{1}{3}\right)$

The coordinates of the foot of perpendiculars from the point (2,3) on the line y=3x+4 is given by (A) ((37)/(1

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(2,3)

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Concept understanding — Distance From Point To Line

Distance from a Point to a Line

The distance from a point to a line is the shortest distance — the length of the perpendicular dropped from the point onto the line. In 3D we compute it with vectors and the cross product.

Let the line be $\vec{r}=\vec{a}+\lambda\vec{b}$ (a point $A$ with position vector $\vec{a}$ , direction $\vec{b}$ ), and let $P$ be the given point with position vector $\vec{p}$ .

The idea

Look at the triangle formed by $A$ , $P$ and the foot of the perpendicular $M$ . The segment $\overrightarrow{AP}=\vec{p}-\vec{a}$ is the hypotenuse, and the perpendicular distance $d=PM$ is the side opposite the angle $\theta$ between $\overrightarrow{AP}$ and the line:

d = |\overrightarrow{AP}|\sin\theta.

But the cross product already contains $\sin\theta$ : $|\overrightarrow{AP}\times\vec{b}| = |\overrightarrow{AP}|\,|\vec{b}|\sin\theta$ . Dividing by $|\vec{b}|$ isolates the distance.

d = \frac{|(\vec{p}-\vec{a})\times\vec{b}|}{|\vec{b}|}

Example

Distance of $P(1,2,3)$ from the line $\vec{r}=(\hat{i}+\hat{j})+\lambda(2\hat{i}-\hat{j}+2\hat{k})$ .

Here $\vec{a}=(1,1,0)$ , $\vec{b}=(2,-1,2)$ , and $\overrightarrow{AP}=\vec{p}-\vec{a}=(0,1,3)$ .

$\overrightarrow{AP}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&1&3\\2&-1&2\end{vmatrix} = (2+3)\hat{i} - (0-6)\hat{j} + (0-2)\hat{k} = 5\hat{i}+6\hat{j}-2\hat{k}.$ …

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The foot of the perpendicular from a point to a line is found by solving the intersection of the given line with the line through the point that has slope equal to the negative reciprocal of the given line’s slope. For point $(2,3)$ and line $y=3x+4$ , the foot is $\left(-\frac{1}{10},\frac{37}{10}\right)$ , which matches option (B).

The core idea: the shortest distance from a point to a line is along the perpendicular. So the foot of the perpendicular is simply the point where the perpendicular from the given point meets the given line. To find it, we need two things: the equation of the perpendicular line, and then its intersection with the original line.

Why the negative reciprocal?

If a line has slope $m$ , any line perpendicular to it has slope $-\frac{1}{m}$ (provided $m \neq 0$ ). This is because the product of slopes of perpendicular lines is $-1$ .

Here, the given line is $y = 3x + 4$ , so its slope is $m = 3$ .

Therefore, the perpendicular slope is $m_\perp = -\frac{1}{3}$ .

Now let’s work through the steps.

Equation of the perpendicular line through $(2,3)$ Using point-slope form:

y - 3 = -\frac{1}{3}(x - 2)

Multiply through by 3:

3y - 9 = -x + 2

Rearranging:

x + 3y = 11 \quad \text{(Equation of perpendicular)}

Find the intersection of this perpendicular with the original line The original line is:

y = 3x + 4

Substitute this into $x + 3y = 11$ :

x + 3(3x + 4) = 11

x + 9x + 12 = 11

10x = -1

x = -\frac{1}{10}

Find the corresponding $y$ coordinate Using $y = 3x + 4$ : $y = 3\left(-\frac{1}{10}\right) + 4 = -\frac{3}{10} + 4 = \frac{37}{10}$ …

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