Answer :

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)

Rate this question :

The equation of aTamil Nadu Board - Math

The centre of a cTamil Nadu Board - Math

If the line segmeTamil Nadu Board - Math

If (1, 2)</Tamil Nadu Board - Math

Find the equationTamil Nadu Board - Math

The point PTamil Nadu Board - Math

The midpoint of tTamil Nadu Board - Math

Find the points wTamil Nadu Board - Math

Find the coordinaTamil Nadu Board - Math

Find the ratio inTamil Nadu Board - Math