Question

The line L: $6x+3y+k=0$ divides the line segment joining the points (3,5) and (4,6) in the ratio -5:4. If the point of intersection of the lines L = 0 and $x-y+1=0$ is P(g,h) then h =

• A. 2g
• B. 2g-1
• C. 3g
• D. g+1

Hint:

A ratio of $-m:n$ implies external division in the ratio $m:n$. The formula for this is $\left(\frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}\right)$. After finding the equation of a line, solving a system of two linear equations can be done quickly by either substitution or elimination.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
This problem has two parts. First, we use the section formula to find the point where line L divides the given segment. Since this point lies on L, we can find the value of $k$. Second, we find the intersection point of line L and another given line and then check the relationship between its coordinates.
Step 2: Key Formula or Approach
1. Use the section formula for a point dividing a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m:n$. Since the ratio is negative, it's external division. Let the ratio be $m:n = 5:(-4)$. The point of division is $P = \left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)$. 2. Substitute the coordinates of this point into the equation of line L to find $k$. 3. Solve the system of linear equations for L=0 and $x-y+1=0$ to find the intersection point $(g,h)$. 4. Check which of the given options correctly relates $g$ and $h$.
Step 3: Detailed Explanation
1. Find k: The points are A(3,5) and B(4,6). The line L divides AB in the ratio -5:4. This means the division point P divides the segment externally in the ratio 5:4. Let's use the external division formula $P = \frac{mB-nA}{m-n}$. \[ x = \frac{5(4) - 4(3)}{5-4} = \frac{20-12}{1} = 8 \] \[ y = \frac{5(6) - 4(5)}{5-4} = \frac{30-20}{1} = 10 \] The point of division is $(8,10)$. This point lies on the line L: $6x+3y+k=0$. Substitute the point into the equation: \[ 6(8) + 3(10) + k = 0 \] \[ 48 + 30 + k = 0 \implies 78 + k = 0 \implies k = -78 \] So, the equation of line L is $6x+3y-78=0$, which simplifies to $2x+y-26=0$. 2. Find the intersection point P(g,h): We need to solve the system of equations: \[ 2x+y-26 = 0 \quad \text{(Line L)} \] \[ x-y+1 = 0 \quad \text{(Second Line)} \] Adding the two equations to eliminate $y$: \[ (2x+y-26) + (x-y+1) = 0 \] \[ 3x - 25 = 0 \implies x = \frac{25}{3} \] So, $g = \frac{25}{3}$. Substitute the value of $x$ back into the second line's equation to find $y$: \[ \frac{25}{3} - y + 1 = 0 \implies y = \frac{25}{3} + 1 = \frac{25+3}{3} = \frac{28}{3} \] So, $h = \frac{28}{3}$. 3. Relate h and g: We have $g = 25/3$ and $h = 28/3$. We check the options: (A) $2g = 2(25/3) = 50/3 \neq h$. (B) $2g-1 = 50/3 - 1 = 47/3 \neq h$. (C) $3g = 3(25/3) = 25 \neq h$. (D) $g+1 = 25/3 + 1 = 28/3 = h$. This is correct. Step 4: Final Answer
The relation between the coordinates of the intersection point is $h=g+1$.