Find the equation of the straight line segment whose end points are the point of intersection of the straight lines

2x – 3y + 4 = 0, x – 2y + 3 = 0 and the midpoint of the line joining the points (3, –2) and (–5, 8).

Here its given that the straight line segment has end points as the point of intersection of the straight lines

2x – 3y + 4 = 0, x – 2y + 3 = 0 and the midpoint of the line joining the points (3, –2) and (–5, 8).

As one end point is point of intersection of the straight lines

2x – 3y + 4 = 0, x – 2y + 3 = 0,we will solve these equations.

2x–3y + 4 = 0––––(1)

x–2y + 3 = 0–––––(2)

Multiply (2) by 2,then we have

i.e. y = 2

Substitute y = 2 in 2x–3y + 4 = 0

2x–3(2) + 4 = 0

⇒ 2x–6 + 4 = 0

⇒ 2x–2 = 0

⇒ 2x = 2

⇒ x = 1

Therefore the lines intersect at (1,2).

Now the line has one end point as (1,2) and other end point as mid– point of the line joining (3, –2) and (–5, 8).

The mid– point of (3, –2) and (–5, 8) is

=

=

= (–1,3)

Hence to find the equation of the line with end points as (x1,y1) and (x2,y2) ,we use:

,where (x1,y1) and (x2,y2) are (1,2) and (–1,3)

⇒

⇒

⇒ –2(y–2) = (x–1)

⇒ –2y + 4 = x–1

⇒ –2y + 4–x + 1 = 0

⇒ –2y–x + 5 = 0

⇒ x + 2y–5 = 0(multiply by –1 on both the sides of the equation)