Question

The straight lines l₁ and l₂ pass through the origin and trisect the line segment of the line L: 9x + 5y = 45 between the axes. If m₁ and m₂ are the slopes of the lines l₁ and l_2, then the point of intersection of the line y = (m₁ + m₂)x with L lies on

• A. y-2x=5
• B. 6x-y=15
• C. y-x=5
• D. 6x+y=10

Answer 1

The Correct Option is C

Line Trisection Geometry Problem 

The line L : 9x + 5y = 45 intersects the axes at A(5, 0) and B(0, 9). The lines l1 and l2 trisect the segment AB.

Let’s find the coordinates of the points P and Q that trisect AB:

• Point P divides AB in the ratio 1:2. Using the section formula:

\( x_P = \frac{1(0) + 2(5)}{1 + 2} = \frac{10}{3} \)

\( y_P = \frac{1(9) + 2(0)}{1 + 2} = 3 \)

So, P = \( \left(\frac{10}{3}, 3\right) \)

• Point Q divides AB in the ratio 2:1. Using the section formula:

\( x_Q = \frac{2(0) + 1(5)}{1 + 2} = \frac{5}{3} \)

\( y_Q = \frac{2(9) + 1(0)}{1 + 2} = 6 \)

So, Q = \( \left(\frac{5}{3}, 6\right) \)

Now, let’s find the slopes m1 and m2:

\( m_1 = \frac{3 - 0}{\frac{10}{3} - 0} = \frac{9}{10} \)

\( m_2 = \frac{6 - 0}{\frac{5}{3} - 0} = \frac{18}{5} \)

Therefore:

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

The equation of the line passing through the origin with slope \( \frac{9}{2} \) is: \( y = \frac{9}{2}x \) or \( 9x - 2y = 0 \).

Now we find the intersection of this line with line L : 9x + 5y = 45:

We have the system:

\( 9x + 5y = 45 \)

\( 9x - 2y = 0 \)

Subtracting the second equation from the first: \( 7y = 45 \), so \( y = \frac{45}{7} \).

Substituting this value into \( 9x - 2y = 0 \): \( 9x = 2\left(\frac{45}{7}\right) = \frac{90}{7} \), so \( x = \frac{10}{7} \).

The point of intersection is \( \left(\frac{10}{7}, \frac{45}{7}\right) \). Let’s check which equation this point satisfies:

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

Conclusion:

Therefore, the point of intersection lies on the line \( \mathbf{y - x = 5} \).