The ratio in which the line 3x+4y+2=0 divides the distance between the lines 3x+4y+5=0 and 3x+4y-5=0 is (A) 1:

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3x+4y+2=0

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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 $3x+4y+2=0$ lies between two parallel lines; we find its perpendicular distances from each and take their ratio. The answer is $3:7$ .

Why this approach works

All three lines are parallel (same coefficients of $x$ and $y$ ), so they form a family of parallel lines in the plane. The "distance between" two parallel lines means the perpendicular distance separating them. When a third parallel line lies between the other two, it divides this perpendicular separation into two segments. The ratio we seek is simply the ratio of these two perpendicular distances.

The perpendicular distance from a point $(x_0, y_0)$ to a line $ax+by+c=0$ is given by

d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2+b^2}}.

For parallel lines of the form $ax+by+c_1=0$ and $ax+by+c_2=0$ , the distance between them is

d = \frac{|c_1 - c_2|}{\sqrt{a^2+b^2}}.

\text{Distance between } ax+by+c_1=0 \text{ and } ax+by+c_2=0 \text{ is } \frac{|c_1-c_2|}{\sqrt{a^2+b^2}}.

Step-by-step solution

1. Identify the structure

The three lines are:

$L_1: 3x+4y+5=0$
$L: 3x+4y+2=0$ (the dividing line)
$L_2: 3x+4y-5=0$

All have the form $3x+4y+k=0$ , so they are parallel. The constant terms are $+5$ , $+2$ , and $-5$ respectively. Since $-5 < 2 < 5$ , the line $L$ lies between $L_1$ and $L_2$ .

2. Compute the distance from $L$ to $L_1$

Using the formula for distance between parallel lines: …

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Q.The ratio in which the line 3x+4y+2=03x+4y+2=03x+4y+2=0 divides the distance between the lines 3x+4y+5=03x+4y+5=03x+4y+5=0 and 3x+4y−5=03x+4y-5=03x+4y−5=0 is (A) 1:21:21:2 (B) 3:73:73:7 (C) 2:32:32:3 (D) 2:52:52:5

Why this approach works

Step-by-step solution

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Q.The ratio in which the line $3x+4y+2=0$ divides the distance between the lines $3x+4y+5=0$ and $3x+4y-5=0$ is
(A) $1:2$
(B) $3:7$
(C) $2:3$
(D) $2:5$

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