A straight line through the origin O meets the parallel lines 4x + 2y = 9 and 2x + y + 6 = 0 at poin

Q: Wait, why can't we just find the intersection points P and Q and use the distance formula?

You absolutely can! That is a valid, brute-force method. However, in JEE Advanced, time is your most precious resource. Finding the intersection of a line through the origin $y = mx$ with two other lines involves solving systems of equations, which is prone to calculation errors. The geometric approach using similar triangles is elegant, faster, and significantly reduces the chance of making a silly mistake.

Q: Does the origin have to be the point of division?

Yes, the problem specifically states that the line passes through the origin $O(0,0)$ and that this point $O$ divides the segment $PQ$. If the line passed through a different point, the ratio would change, but the geometric principle of similar triangles would still apply, just with a different reference point for the perpendicular distances.

Q: Why do we use the absolute value in the distance formula?

The perpendicular distance formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$ uses the absolute value because distance must always be a positive quantity. The expression $ax_0 + by_0 + c$ can be negative depending on which side of the line the point lies, but the physical distance is the magnitude of that value. Since we are looking for a ratio of distances, keeping them positive is essential.

Question

A straight line through the origin $O$ meets the parallel lines $4x + 2y = 9$ and $2x + y + 6 = 0$ at points $P$ and $Q$ respectively. Then the point $O$ divides the segment $PQ$ in the ratio

Options:

(A) 1:2

(B) 3:4

(C) 2:1

(D) 4:3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to look at a problem that, at first glance, might seem like a standard coordinate geometry exercise.

You are given two lines, $4x + 2y = 9$ and $2x + y + 6 = 0$, and a transversal line passing through the origin $O(0,0)$ that intersects them at $P$ and $Q$. The question asks for the ratio in which the origin divides the segment $PQ$.

Many students would immediately jump into solving for the intersection points $P$ and $Q$. While that works, it is a path filled with algebraic traps. Let us instead embrace the beauty of geometry.

### The Parallel Trap

First, let us look at our lines. The first line is $L_1: 4x + 2y - 9 = 0$. The second is $L_2: 2x + y + 6 = 0$.

If you calculate the slope of $L_1$, you get $m_1 = -\frac{4}{2} = -2$. For $L_2$, you get $m_2 = -\frac{2}{1} = -2$.

They are parallel! This is not a coincidence; it is a gift. When you have parallel lines, the geometry becomes much simpler.

Imagine the origin $O$ as a pivot point. The line passing through it acts as a transversal. Because the lines are parallel, the triangles formed by the origin and the segments on the lines are similar. This is the core geometric reality you must visualize.

### The Geometric Shortcut

Because the triangles are similar, the ratio of the segments $OP$ and $OQ$ is directly proportional to the perpendicular distances from the origin to the lines.

Specifically, the ratio is given by:

$$\frac{OP}{OQ} = \frac{d_1}{d_2}$$

Here, $d_1$ is the distance from the origin to $L_1$, and $d_2$ is the distance from the origin to $L_2$. By using this property, we completely bypass the need to find the coordinates of $P$ and $Q$. We are no longer doing heavy algebra; we are doing elegant geometry.

### The Calculation

Now, we apply the perpendicular distance formula:

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

For $L_1$, with the origin $(0,0)$, we have:

$$d_1 = \frac{|4(0) + 2(0) - 9|}{\sqrt{4^2 + 2^2}} = \frac{|-9|}{\sqrt{16 + 4}} = \frac{9}{\sqrt{20}}$$

We can simplify $\sqrt{20}$ to $2\sqrt{5}$, so $d_1 = \frac{9}{2\sqrt{5}}$. For $L_2$, we have:

$$d_2 = \frac{|2(0) + 1(0) + 6|}{\sqrt{2^2 + 1^2}} = \frac{|6|}{\sqrt{4 + 1}} = \frac{6}{\sqrt{5}}$$

Now, we find the ratio:

$$\frac{OP}{OQ} = \frac{d_1}{d_2} = \frac{\frac{9}{2\sqrt{5}}}{\frac{6}{\sqrt{5}}}$$

The $\sqrt{5}$ terms cancel out beautifully, leaving us with:

$$\frac{9}{2} 	imes \frac{1}{6} = \frac{9}{12} = \frac{3}{4}$$

The origin divides the segment $PQ$ in the ratio **$3:4$**. This is the power of looking for the geometric soul of a problem. You did not need to solve for $x$ or $y$; you only needed to understand the relationship between the lines.

Animated Solution for Mathematics - Straight Lines: A straight line through the origin $O$ meets the parallel lines $4 x + 2 y = 9$ and $2 x + y + 6 = 0$ at points $P$ and $Q$ respectively. Then the point $O$ divides the segment $P Q$ in the ratio

Select Answer:

Visualized Solution

Visualizing the Given Lines

Identifying Parallel Lines

Drawing the Transversal

The Geometric Property

Visualizing Perpendicular Distances

Perpendicular Distance Formula

Setting up $d_{1}$

Calculating $d_{1}$

Setting up $d_{2}$

Calculating $d_{2}$

Setting up the Ratio

Simplifying the Ratio

Final Conclusion

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

The Parallel Trap

The Geometric Shortcut

The Calculation

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