Find the direction in which a straight line must be drawn through the point $(-1, 2)$ so that its point of intersection with the line $x + y = 4$ may be at a distance of $3$ units from this point.

Find the direction in which a straight line must be drawn through the point (-1, 2) so that its point of inter

Miscellaneous Exercise · Q15

(-1, 2)

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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 line through $(-1, 2)$ must meet $x + y = 4$ at a point $3$ units away. The two such points are $(2, 2)$ and $(-1, 5)$ , giving a horizontal direction (slope $0$ , $\theta = 0^\circ$ ) and a vertical direction (undefined slope, $\theta = 90^\circ$ ).

Step-by-step solution

1. Parametrise the line through $P(-1, 2)$ . A point at distance $r$ along a direction making angle $\theta$ with the $x$ -axis is:

x = -1 + r\cos\theta, \qquad y = 2 + r\sin\theta

2. Require the meeting point on $x + y = 4$ at $r = 3$ . Substituting:

(-1 + 3\cos\theta) + (2 + 3\sin\theta) = 4 \implies 3(\cos\theta + \sin\theta) = 3

\cos\theta + \sin\theta = 1

3. Solve. Writing $\cos\theta + \sin\theta = \sqrt{2}\sin\!\left(\theta + 45^\circ\right)$ :

$\sqrt{2}\sin\!\left(\theta + 45^\circ\right) = 1 \implies \sin\!\left(\theta + 45^\circ\right) = \frac{1}{\sqrt{2}}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Higher Secondary (+2 Stage) Examination 2025Set ANNUAL5 marks
$(1,0,0)$
›Reveal solutionSolution

Write a general point on the line, force the vector from the given point to it to be perpendicular to the line's direction, solve for the parameter, then read off the foot, distance, and perpendicular's equation.

The line is $\dfrac{x-1}{2}=\dfrac{y+1}{-3}=\dfrac{z+10}{8}=t$ , passing through $(1,-1,-10)$ with direction ratios $(2,-3,8)$ . A general point on it is

$Q=(1+2t,\,-1-3t,\,-10+8t).$

The vector from the given point $P(1,0,0)$ to $Q$ is

$\vec{PQ}=(2t,\;-1-3t,\;-10+8t).$

For $Q$ to be the foot of the perpendicular, $\vec{PQ}$ must be perpendicular to the line's direction $(2,-3,8)$ , i.e. their dot product is zero:

$2(2t)+(-3)(-1-3t)+8(-10+8t)=0$

$4t+3+9t-80+64t=0$

$77t-77=0\ \Rightarrow\ t=1.$

Foot of perpendicular: substituting $t=1$ : $Q=(1+2,\,-1-3,\,-10+8)=(3,-4,-2)$ .

Distance: $\vec{PQ}=(3-1,\,-4-0,\,-2-0)=(2,-4,-2)$ , so

$|PQ|=\sqrt{2^2+(-4)^2+(-2)^2}=\sqrt{4+16+4}=\sqrt{24}=2\sqrt6.$

…

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Higher Secondary (+2 Stage) Examination 2023Set ANNUAL5 marks
$(1,6,3)$
›Reveal solutionSolution

Find the foot of the perpendicular from the point to the line (by making the vector from a general point on the line to the given point perpendicular to the line's direction), then the image is the point such that this foot is the midpoint of the given point and its image.

Line: $\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3}=t$ , so a general point on the line is $Q(t)=(t,\,1+2t,\,2+3t)$ , with direction ratios $(1,2,3)$ .

Let $P=(1,6,3)$ .

Find the foot of the perpendicular $F$ : The vector $\overrightarrow{QP} = (1-t,\ 6-(1+2t),\ 3-(2+3t)) = (1-t,\ 5-2t,\ 1-3t)$ must be perpendicular to the direction $(1,2,3)$ :

$(1-t)(1)+(5-2t)(2)+(1-3t)(3)=0$

$1-t+10-4t+3-9t=0$

$14-14t=0 \;\Rightarrow\; t=1$

So the foot of the perpendicular is $F=Q(1)=(1,\,1+2,\,2+3)=(1,3,5)$ .

Find the image $P'$ : Since $F$ is the midpoint of $P$ and its image $P'$ :

$F=\dfrac{P+P'}{2} \;\Rightarrow\; P'=2F-P$

$P' = (2(1)-1,\ 2(3)-6,\ 2(5)-3) = (1,\,0,\,7)$

…

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Q.Find the direction in which a straight line must be drawn through the point (−1,2)(-1, 2)(−1,2) so that its point of intersection with the line x+y=4x + y = 4x+y=4 may be at a distance of 333 units from this point.

Step-by-step solution

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Q.Find the direction in which a straight line must be drawn through the point $(-1, 2)$ so that its point of intersection with the line $x + y = 4$ may be at a distance of $3$ units from this point.

Unlock everything free for 14 days