Question

A line is such that its segment between the lines 5x - y + 4 = 0 and 3x + 4y -4 = 0 is bisected at the point (1,5). Obtain its equation.​

Answer 1

ssume the required line to be: y=mx+c and determine its intersections with the given lines. These are as follows: x=(4+c)/(5-m) & y=(4m+5c)/(5-m) and x=(4)(1-c)/(4m+3) & y= (3c+4m)/(4m+3). Now we set up equations with given midpoint viz. (1,5). We obtain: (4+c)/(5-m) + (4)(1-c)/(4m+3) =2 —-————(1) and y=(4m+5c)/(5-m) + (3c+4m)/(4m+3) =10 —————(2). Solve (1) & (2) to get m= 83/35 & c= 92/35 so that the required line is: y= 83x/35 + 92/35 or 83x- 35y+92=0

Sanity Check: Intersection of 83x- 35y+92=0 & 5x-y-4=0 is P: [58/23,198/23] while Intersection of 83x- 35y+92=0 & 3x+4y-4=0 is Q: [-12/23, 32/23]. It can easily be verified that midpoint of PQ is : (1,5)

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