The straight lines l1 and l2 pass through the origin and trisect the line segment of the line L : 9x

Q: Wait, why do we use the ratio $1:2$ and $2:1$ for trisection?

Trisection means dividing a segment into three equal parts. If you have a segment $AB$, the two points $C$ and $D$ that divide it into three equal parts will be at distances of $1/3$ and $2/3$ of the total length from $A$. Thus, $C$ divides $AB$ in the ratio $1:2$, and $D$ divides it in the ratio $2:1$. It is a beautiful application of the section formula!

Q: Is there a faster way to find the intersection without calculating $m_1$ and $m_2$ separately?

While you could potentially use properties of the centroid or other geometric symmetries, calculating the slopes $m_1$ and $m_2$ is the most robust and foolproof method for JEE Advanced. It minimizes the risk of conceptual errors and ensures you have the exact coordinates for the final verification.

Q: Why do we need to check the options at the end?

In competitive exams like JEE, the final answer might be represented in various forms. Checking the options is a crucial step to ensure your derived intersection point $P$ satisfies the equation of the line given in the correct option. It is the final confirmation that your journey through the problem was correct.

Question

The straight lines $l_1$ and $l_2$ pass through the origin and trisect the line segment of the line $L : 9x + 5y = 45$ between the axes. If $m_1$ and $m_2$ are the slopes of the lines $l_1$ and $l_2$, then the point of intersection of the line $y = (m_1 + m_2)x$ with $L$ lies on

Options:

(A) $y - 2x = 5$

(B) $6x + y = 10$

(C) $y - x = 5$

(D) $6x - y = 15$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a coordinate plane. You have a line $L$ defined by the equation $9x + 5y = 45$.

To truly understand it, we transform it into the intercept form by dividing the entire equation by $45$. This yields:

$$ \frac{x}{5} + \frac{y}{9} = 1 $$

Suddenly, the geometry becomes clear. The line cuts the x-axis at $A(5, 0)$ and the y-axis at $B(0, 9)$. We have our segment $AB$ anchored firmly on the axes.

### The Art of Trisection

The problem asks us to consider two lines, $l_1$ and $l_2$, that pass through the origin and trisect this segment $AB$. Trisection is the act of dividing a whole into three equal parts.

To achieve this, we need two points, $C$ and $D$, that sit on $AB$. Point $C$ divides $AB$ in the ratio $1:2$, and point $D$ divides it in the ratio $2:1$.

We invoke the **Section Formula**:

$$ P = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) $$

For point $C$, with ratio $1:2$, the coordinates become:

$$ x_C = \frac{1(0) + 2(5)}{1+2} = \frac{10}{3}, \quad y_C = \frac{1(9) + 2(0)}{1+2} = 3 $$

Thus, $C$ is at $(\frac{10}{3}, 3)$. For point $D$, with ratio $2:1$, the coordinates are:

$$ x_D = \frac{2(0) + 1(5)}{2+1} = \frac{5}{3}, \quad y_D = \frac{2(9) + 1(0)}{2+1} = 6 $$

Thus, $D$ is at $(\frac{5}{3}, 6)$.

### The Slopes of the Lines

Now, we define our lines $l_1$ and $l_2$. They pass through the origin $(0,0)$ and the points $C$ and $D$ respectively.

The slope $m_1$ of line $l_1$ is:

$$ m_1 = \frac{3 - 0}{10/3 - 0} = \frac{9}{10} $$

The slope $m_2$ of line $l_2$ is:

$$ m_2 = \frac{6 - 0}{5/3 - 0} = \frac{18}{5} $$

We are looking for the intersection of the line $y = (m_1 + m_2)x$ with our original line $L$. First, we sum the slopes:

$$ m_1 + m_2 = \frac{9}{10} + \frac{18}{5} = \frac{9 + 36}{10} = \frac{45}{10} = \frac{9}{2} $$

Our new line is $y = \frac{9}{2}x$.

### The Climax: Finding the Intersection

We now solve the system of equations: $9x + 5y = 45$ and $y = \frac{9}{2}x$. Substituting the second into the first, we get:

$$ 9x + 5\left( \frac{9}{2}x \right) = 45 $$

Multiplying by $2$ to clear the denominator, we have $18x + 45x = 90$, which simplifies to $63x = 90$. Solving for $x$, we find:

$$ x = \frac{90}{63} = \frac{10}{7} $$

Substituting this back into $y = \frac{9}{2}x$, we get:

$$ y = \frac{9}{2} \cdot \frac{10}{7} = \frac{45}{7} $$

The intersection point $P$ is $(\frac{10}{7}, \frac{45}{7})$.

### Final Verification

Finally, we test this point against the condition $y - x = 5$. We calculate:

$$ \frac{45}{7} - \frac{10}{7} = \frac{35}{7} = 5 $$

The equation holds true! You have successfully navigated the trisection and found the intersection. This is the elegance of coordinate geometry.

Select Answer:

Visualized Solution

Visualizing the Line $L$

The Concept of Trisection

Calculating Trisection Point $C$

Calculating Trisection Point $D$

Finding Slopes $m_{1}$ and $m_{2}$

The New Line Equation

Finding the Intersection Point $P$ (X-coordinate)

Finding the Intersection Point $P$ (Y-coordinate)

Verifying the Options

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Art of Trisection

The Slopes of the Lines

The Climax

Final Verification

Similar Questions

A straight line $L$ is perpendicular to the line $5 x - y = 1$ . The area of the triangle formed by the line $L$ and the coordinate axes is 5. Find the equation of the line $L$ .

A line through $A (- 5, - 4)$ meets the line $x + 3 y + 2 = 0$ , $2 x + y + 4 = 0$ and $x - y - 5 = 0$ at the points $B, C$ and $D$ respectively. If $(15/ A B)^{2} + (10/ A C)^{2} = (6/ A D)^{2}$ , find the equation of the line.

The locus of the mid-point of the perpendiculars drawn from points on the line, $x = 2 y$ to the line $x = y$ is:

The equations of two sides $A B$ and $A C$ of a triangle $A B C$ are $4 x + y = 14$ and $3 x - 2 y = 5$ , respectively. The point $(2, - 4/3)$ divides the third side $B C$ internally in the ratio $2 : 1$ . The equation of the side $B C$ is

Let $a, b, c$ and $d$ be non-zero numbers. If the point of intersection of the lines $4 a x + 2 a y + c = 0$ and $5 b x + 2 b y + d = 0$ lies in the fourth quadrant and is equidistant from the two axes then

The locus of mid points of the perpendiculars drawn from points on the line $x = 2 y$ to the line $x = y$ is :

Let $P S$ be the median of the triangle with vertices $P (2, 2), Q (6, - 1)$ and $R (7, 3)$ . The equation of the line passing through $(1, - 1)$ and parallel to $P S$ is

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. The equation of the bisector of the angle $P QR$ is

If $C$ be the centroid of the triangle having vertices $(3, - 1), (1, 3)$ and $(2, 4)$ . Let $P$ be the point of intersection of the lines $x + 3 y - 1 = 0$ and $3 x - y + 1 = 0$ , then the line passing through the points $C$ and $P$ also passes through the point:

Select Answer:

Visualized Solution

Visualizing the Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

The Concept of Trisection

Calculating Trisection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Calculating Trisection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Finding Slopes .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m2​