Find the points on the line x+y=4 which lie at a unit distance from the line 4x+3y=10.

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x+y=4

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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 problem asks for points on line $L_1: x+y=4$ that are exactly 1 unit away from line $L_2: 4x+3y=10$ . Using the perpendicular distance formula, we set up an equation and solve for the intersection parameter. The two points are $(3,1)$ and $(-7,11)$ .

Concept and Intuition

The distance from a point to a line is always measured along the perpendicular. So for any point on $x+y=4$ , we can compute its perpendicular distance to $4x+3y=10$ using the standard formula. The trick is that the point isn't fixed — it slides along the first line. We need to find which positions on that line give exactly a unit distance.

Think of it this way: the line $x+y=4$ is a straight path. As you walk along it, your perpendicular distance to the other line changes continuously. At some points you're closer, at others farther. We're looking for the spots where that distance is exactly 1.

Step-by-Step Solution

1. Parameterize the first line

The line $x+y=4$ can be written as $y = 4 - x$ . Any point on it has coordinates $(t, 4-t)$ where $t$ is a real parameter. This is convenient because we can plug it directly into the distance formula.

2. Recall the distance formula

(x_1, y_1)

For the line $4x + 3y = 10$ , rewrite it as $4x + 3y - 10 = 0$ . So $A = 4$ , $B = 3$ , $C = -10$ .

3. Apply the formula to our point

Plug $(t, 4-t)$ into the numerator:

|4(t) + 3(4-t) - 10| = |4t + 12 - 3t - 10| = |t + 2|

The denominator is $\sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ .

So the distance from $(t, 4-t)$ to the line $4x+3y=10$ is:

d = \frac{|t + 2|}{5}

4. Set the distance equal to 1

We want $d = 1$ , so:

$\frac{|t + 2|}{5} = 1 \quad \Rightarrow \quad |t + 2| = 5$ …

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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 points on the line x+y=4x+y=4x+y=4 which lie at a unit distance from the line 4x+3y=104x+3y=104x+3y=10.

Concept and Intuition

Step-by-Step Solution

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Q.Find the points on the line $x+y=4$ which lie at a unit distance from the line $4x+3y=10$ .

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