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

y = 1/3x − 3

y = 1/3x + 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 = 1/3x − 3

 y = 1/3x + 17

 y = −3x − 3

 y = −3x + 17

johnweldon1993 · Answer

we have a point....(6,-1)
we have an equation....and we want a perpendicular line right?

a perpendicular line has the negative reciprocal slope of the other line ...so

3x + y = 7  in slope intercept form is

y = -3x + 7

the negative reciprocal of -3 is?

well -3...is the same as -3/1 right? the negative (turns this into positive) reciprocal (flip the fraction) is 1/3x

so

we have a slope now....and a point....we need the point slope formula

y - y1 = m(x-x1)

m is slope
(6, −1) is our point
(x1 , y1)

plug everything in..what do you get?

anonymous · Answer

c?

johnweldon1993 · Answer

y - (-1) = 1/3(x-6)
y + 1 = 1/3(x-6)
         = 1/3x -2
y = 1/3x -3