Question

find the equation to the line passing through the point (6-4) and perpendicular to the line 7x-6y+3=0​

Answer 1

Answer:

6x + 7y - 8 = 0

Step-by-step explanation:

As the that line is perpendicular to 7x - 6y + 3 = 0, slope of that that will be -ve reciprocal of slope of line 7x - 6y + 3 = 0.

For slope of 7x - 6y + 3 = 0 :

         ⇒ 7x + 3 = 6y

         ⇒ y = (7/6)x + (3/6)

So, slope of 7x - 6y + 3 is 7/6, and hence slope of the needed line is - 6/7.

As that line passes through ( 6, - 4 ).

⇒ eq. of that line is :

⇒ y - ( - 4 ) = m( x - 6 )   { m = slope }

⇒ y + 4 = (-6/7)( x - 6 )

⇒ 7( y + 4 ) = -6( x - 6 )

⇒ 7y + 28 = - 6x + 36

⇒ 6x + 7y + 28 - 36 = 0

⇒ 6x + 7y - 8 = 0

     Required eq. is 6x + 7y - 8 = 0