Question

Find the equations of this lines through the intersection of x-y-7=0 and x+y=0 and which is perpendicular to 2x-3y=4

Answer 1

GIVEN.

equation of lines through which intersection

= x - y - 7 = 0 and x + y = 0

perpendicular to the line = 2x - 3y = 4

=> we can solve the equation of intersection

=> x - y = 7 .....(1)

=> x + y = 0 .... (2)

= we get,

=> 2x = 7

=> x = 7/2

put the value of x = 7/2 in equation (2)

we get,

=> 7/2 + y = 0

=> y = -7/2

Therefore,

point are = [ 7/2 , -7/2 ]

equation of perpendicular = 2x - 3y = 4

=> we can find slope of equation

=> slope of line perpendicular to the line

= b/a

=> 2x - 3y - 4 = 0

=> 3/2 = slope

equation of line = ( y - Y1) = m( x - X1)

=> ( y + 7/2 ) = 3/2 ( x - 7/2 )

=> ( 2y + 7 / 2 ) = 3/2 ( 2x - 7 / 2 )

=> ( 2y + 7 / 2 ) = ( 6x - 21 / 4 )

=> 2 ( 2y + 7 ) = 6x - 21

=> 4y + 14 = 6x - 21

=> 6x - 4y - 21 - 14 = 0

=> 6x - 4y = 35

Therefore,

equation of lines = 6x - 4y = 35.