Question

Find the equation of a straight line joining the point intersection of
3x + y +2 = 0 and x - 2y - 4 = 0 to the
point of intersection
7x – 3y = - 12
and 2y = x + 3.​

Answer 1

Answer:

For the point of intersection of 3x + y + 2 = 0 and x - 2y - 4 = 0 -

3x + y + 2 = 0

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

= > x - 2y - 4 = 0

= > x - 2( - 2 - 3x ) - 4 = 0

= > x + 4 + 6x - 4 = 0

= > 7x = 0

= > x = 0

Hence,

= > y = - 2 - 3(0)

= > y = - 2

Point of intersection is ( 0, -2 )

For the point of intersection of 7x - 3y = - 12 and 2y = x + 3 -

= > 7x - 3y = - 12

= > (7/3)x - y = - 4

= > (7/3)x + 4 = y

Hence,

= > 2y = x + 3

= > 2[ (7/3)x + 4 ] = x + 3

= > (14/3)x + 8 = x + 3

= > (11/3)x = - 5

= > x = - 15/11

Thus,

= > 2y = x + 3

= > y = ( - 15/11 + 3 )/2

= > y = ( - 15 + 33 ) / 22

= > y = 9/11

Point of intersection = (-15/11 , 9/11 )

Hence,

Slope of the line so formed -

= > [ 9/11 + 2 ] / [ -15/11 - 0 ]

= > ( 31 / 11 ) / ( - 15/11 )

= > - 31/15

Hence, equ. for that line is -

= > ( x - 0 ) = (-31/15)( y + 2 )

= > 15x = - 31y - 62

= > 15x + 31y + 62 = 0