Question

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

Answer 1

To find the equation of the line, we need its slope, and a point on that line.
We already have a point (6, -1)
We need its slope...

So what is the slope of this line?

Answer 2

two lines are perpendicular to each other if m1=-1/m2...where m1 and m2 are slopes

Answer 3

here slope of this line=-3...

Answer 4

so slope of our line will be -1/-3=1/3

Answer 5

now use point slope form

Answer 6

y-y0=m(x-x0) ---->y-6=1/3(x+1)

Answer 7

-x+3y=19

Answer 8

okay?

Answer 9

got it @AlmightySosa300 ?

Answer 10

??

Answer 11

eq of given line is line 3x + y = 7. It can also be expressed as
y = -3x+7 comparing it with y=mx+c we find slope m= -3
therefore the slope of the line perpendicular to it i.e. \[m=\frac{1}{3}\]since required line passes through the point (6, −1).
threfore \[x_{1}=6 \]
\[y_{1}=-1 \]
eq of required line in slope point form is given as
\[y-y_{1}=m(x-x_{1})\] substituting the respective values in the formula we find
\[y-(-1)=\frac{1}{3}(x-6) \rightarrow y+1=\frac{1}{3}x-2\]
i.e. \[ y=\frac{1}{3}x-3 \rightarrow 3y=x-9 \rightarrow -x+3y=-9 \rightarrow x-3y=9\] which is the required eq of the line

Answer 12

y = one thirdx − 3

y = one thirdx + 17

y = −3x − 3

y = −3x + 17