Question

18. 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).

Answer 1

Hi ,

i ) Let P is the intersecting point of

Straight lines
2x - 3y + 4 = 0 ---( 1 )
x- 2y + 3 = 0
x = 2y - 3 ----( 2 )

Substitute x = 2y - 3 in equation ( 1 ),
we get ,

2(2y - 3) - 3y + 4 = 0
4y - 6 - 3y + 4 = 0
y -2 = 0
y = 2

Put y = 2 in equation ( 2 ) , we get

x = 2 × 2 - 3
x = 1

P( x , y ) = ( 1 , 2 )

ii )
*******************************************
If A( x1 , y1) , B(x2 , y2 ) are any
two points, then the mid point if the
segment joining AB is
Q ( x , y )
x = ( x1 + x2 )/2

y = ( y1 + y2 )/2
***************************************
Let Q( x , y ) is the mid point of

the segment joining A( 3 ,-2 ) , B(-5,8)

x = ( 3 - 5 )/2 = -2/2 = -1

y = ( -2 + 8)/2 = 6/2 = 3

Q( x , y ) = ( -1 , 3 )

iii ) Equation of the line passing
through P( 1 , 2 ) , Q( -1 , 3 )

P( x1 , y1 ) = ( 1 , 2 )
Q( x2 , y2 ) = ( -1 , 3 )

( y - y1 )( x2 - x1 ) = ( x - x1 )( y2 - y1 )

( y - 2 )( -1 - 1 ) = ( x - 1 )( 3 - 2 )

( y - 2 )( -2 ) = ( x - 1 )

-2y + 4 = x - 1

x + 2y - 5 = 0

Therefore ,

Required equation is

x + 2y - 5 = 0

I hope this helps you.

: )