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

1
by Deleted account

• Brainly User
2014-10-05T14:58:20+05:30
Let ab + by +c = 0 be the equation of a line.
This can be written in the form y = -(a/b)x - c
If (X,Y) be a point on above line,
then corresponding to abscissa X, Y will be of the form -(1/b)X - c
That is a point (X,Y) can be written as {X, -(a/b)X - c}.
this fact will be utilised in this question
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Line 5x - y - 4 = 0 can be written as y = 5x + 4 -------(1)

Point (a,5a+4) lies on this line. ---------------(2)
Let this be the point where it meets the line segment with (1,5) as mid-point.

Line 3x + 4y - 4=0 can be written as y = - (3/4) + 4 ----------(2)

Point (4b, -3b+1) lies on this line----------------(3)
Let this be the point where it meets the line segment with (1,5) as mid-point.

(1,5) is the mid-point of points given in (2) and (3),
Note:--mid-point of points (p,q) and (v,w) is [(p+v)/2 , (q+w)/2]
Hence  (a+4b)/2 = 1 ==> a + 4b = 2 ------------------(4)
Also {(5a+4) + (-3b+1)}/2 = 5 ==> 5a - 3b = 5  ---------(5)

Solving (4) and (5) for a and b,
a = 26/23, and b = 5/23     ------------------------(6)

From (3) and (6),
Point of intersection of line (1) and segment = (26/23, 222/23) ---------(7)
Point of intersection of line (2) and segment = (20/23, 8/23) ------------(8)

Equation of line whose segment is a part is given by-
(y -5) = [{(8/23) - 5} / {(20/23) - 1}](x - 1)

This reduces to 107x - 3y = 92