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: Use Ratio Formula: Usually, the ratio \( m:n \) in which a line \( Ax+By+C=0 \) divides the segment joining \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( -\frac{Ax_1+By_1+C}{Ax_2+By_2+C} \). However, here the ratio is given as \( -5:4 \). Let's interpret this as the division point being external or simply find the point that divides the segment in that ratio and lies on the line. Let the point of intersection be \( R \). Using section formula for ratio \( -5:4 \): \( x_R = \frac{-5(4) + 4(3)}{-5+4} = \frac{-20+12}{-1} = 8 \) \( y_R = \frac{-5(6) + 4(5)}{-5+4} = \frac{-30+20}{-1} = 10 \) Point \( R(8,10) \) lies on line \( L \).
Step 2: Find Equation of Line L: Substitute \( (8,10) \) into \( 6x+3y+k=0 \): \( 6(8) + 3(10) + k = 0 \) \( 48 + 30 + k = 0 \implies k = -78 \). So \( L: 6x+3y-78=0 \), which simplifies to \( 2x+y-26=0 \).
Step 3: Find Intersection P(g,h): Solve system: 1) \( 2x + y = 26 \) 2) \( x - y = -1 \) Add equations: \( 3x = 25 \implies g = 25/3 \). Substitute in (2): \( y = x + 1 \implies h = 25/3 + 1 = 28/3 \).
Step 4: Check Options: We need relation between \( h \) and \( g \). From \( y = x + 1 \), we have \( h = g + 1 \). Option (D) matches.