Key Concepts and Formulas

• Intercept Form of a Line: The equation of a line with x-intercept $a$ and y-intercept $b$ is given by $\frac{x}{a} + \frac{y}{b} = 1$.
• Section Formula: The coordinates of a point $P(x, y)$ dividing the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m:n$ are given by $P\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$.
• Slope of a Line: The slope of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$. A line passing through the origin $(0,0)$ and $(x_1, y_1)$ has slope $m = \frac{y_1}{x_1}$.

Step-by-Step Solution

Step 1: Find the x and y intercepts of the line L.

We are given the equation of line L as $9x + 5y = 45$. To find the points where this line intersects the x and y axes, we set $y=0$ and $x=0$ respectively.

• To find the x-intercept, set $y = 0$: $9x + 5(0) = 45$ $9x = 45$ $x = 5$ So, the x-intercept is the point $A(5, 0)$.

• To find the y-intercept, set $x = 0$: $9(0) + 5y = 45$ $5y = 45$ $y = 9$ So, the y-intercept is the point $B(0, 9)$.

We need these intercepts to define the line segment that is being trisected.

Step 2: Find the coordinates of the trisection points P1 and P2.

We are given that lines $l_1$ and $l_2$ trisect the line segment AB. This means we need to find two points, $P_1$ and $P_2$, that divide the segment into three equal parts. Let $P_1$ divide AB in the ratio 1:2 and $P_2$ divide AB in the ratio 2:1.

• To find $P_1$, use the section formula with $m = 1$, $n = 2$, $A(5, 0)$ and $B(0, 9)$: $P_1 = \left(\frac{1(0) + 2(5)}{1+2}, \frac{1(9) + 2(0)}{1+2}\right)$ $P_1 = \left(\frac{10}{3}, \frac{9}{3}\right) = \left(\frac{10}{3}, 3\right)$

• To find $P_2$, use the section formula with $m = 2$, $n = 1$, $A(5, 0)$ and $B(0, 9)$: $P_2 = \left(\frac{2(0) + 1(5)}{2+1}, \frac{2(9) + 1(0)}{2+1}\right)$ $P_2 = \left(\frac{5}{3}, \frac{18}{3}\right) = \left(\frac{5}{3}, 6\right)$

These points $P_1$ and $P_2$ are necessary to determine the slopes of $l_1$ and $l_2$, which pass through the origin.

Step 3: Calculate the slopes m1 and m2 of lines l1 and l2.

Since $l_1$ and $l_2$ pass through the origin (0, 0) and the trisection points $P_1\left(\frac{10}{3}, 3\right)$ and $P_2\left(\frac{5}{3}, 6\right)$ respectively, we can find their slopes using the formula $m = \frac{y}{x}$.

• The slope of $l_1$ is: $m_1 = \frac{3}{\frac{10}{3}} = \frac{3 \cdot 3}{10} = \frac{9}{10}$

• The slope of $l_2$ is: $m_2 = \frac{6}{\frac{5}{3}} = \frac{6 \cdot 3}{5} = \frac{18}{5}$

These slopes are needed to find the equation of the line $y = (m_1 + m_2)x$.

Step 4: Calculate the sum of the slopes m1 + m2.

We need to find the sum $m_1 + m_2$ to define the line $y = (m_1 + m_2)x$. $m_1 + m_2 = \frac{9}{10} + \frac{18}{5} = \frac{9}{10} + \frac{36}{10} = \frac{45}{10} = \frac{9}{2}$

Step 5: Determine the equation of the line y = (m1 + m2)x.

Substituting the value of $m_1 + m_2$ into the equation, we get: $y = \frac{9}{2}x$

This equation is required to find the intersection point with line L.

Step 6: Find the point of intersection of the lines 9x + 5y = 45 and y = (9/2)x.

Substitute $y = \frac{9}{2}x$ into the equation $9x + 5y = 45$: $9x + 5\left(\frac{9}{2}x\right) = 45$ $9x + \frac{45}{2}x = 45$ Multiply by 2 to eliminate the fraction: $18x + 45x = 90$ $63x = 90$ $x = \frac{90}{63} = \frac{10}{7}$

Now, substitute $x = \frac{10}{7}$ back into $y = \frac{9}{2}x$: $y = \frac{9}{2} \cdot \frac{10}{7} = \frac{9 \cdot 5}{7} = \frac{45}{7}$

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

This intersection point is what we'll use to find the line that passes through it.

Step 7: Check which of the given options the point of intersection lies on.

We test each option with the point $\left(\frac{10}{7}, \frac{45}{7}\right)$.

• (A) $6x - y = 15$ $6\left(\frac{10}{7}\right) - \frac{45}{7} = \frac{60}{7} - \frac{45}{7} = \frac{15}{7} \neq 15$

• (B) $6x + y = 10$ $6\left(\frac{10}{7}\right) + \frac{45}{7} = \frac{60}{7} + \frac{45}{7} = \frac{105}{7} = 15 \neq 10$

• (C) $y - x = 5$ $\frac{45}{7} - \frac{10}{7} = \frac{35}{7} = 5$

• (D) $y - 2x = 5$ $\frac{45}{7} - 2\left(\frac{10}{7}\right) = \frac{45}{7} - \frac{20}{7} = \frac{25}{7} \neq 5$

Only option (C) is satisfied by the point $\left(\frac{10}{7}, \frac{45}{7}\right)$.

Common Mistakes & Tips

• Be careful with the section formula; ensure you use the correct ratio (m:n) and the correct coordinates of the endpoints.
• When calculating slopes, remember the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$. If one of the points is the origin, this simplifies to $m = \frac{y}{x}$.
• Double-check arithmetic, especially when working with fractions, to avoid errors.

Summary

We found the x and y intercepts of the line L, then determined the coordinates of the trisection points using the section formula. These points, along with the origin, were used to calculate the slopes of lines $l_1$ and $l_2$. The sum of these slopes defined a new line, whose intersection point with L was found. Finally, we verified that the intersection point lies on the line $y - x = 5$.

The final answer is \boxed{5}, which corresponds to option (C).