Write the equation of the line that is perpendicular to the line 3x + y = 7 and passes through the point (6, −1).

y = one thirdx − 3

y = one thirdx + 17

y = −3x − 3

y = −3x + 17

Question

Write the equation of the line that is perpendicular to the line 3x + y = 7 and passes through the point (6, −1). 

 y = one thirdx − 3 

 y = one thirdx + 17 

 y = −3x − 3 

 y = −3x + 17

anonymous · Answer

If the line is perependicular 
then the product of their slopes is  equal to -1
m1m2=-1
Here the slope of the line is
y=-3x+7
m1=-3
m2=1/3
There fore
the required eqn is
y-6=1/3(x+1)
3y-18=x+1
x-3y+19=0