find the equation...?

Question

of the line parallel to y=-1/3x-7 through (6,5)

Answer 1

m = (- 1 / 3)
y - 5 = (- 1 / 3) (x - 6)
3y - 15 = - x + 6
x + 3y - 21 = 0
OR
3y = - x + 21
y = (- 1 / 3) x + 7

Answer 2

Suppose you're given a line and its slope (m). Any parallel line has a slope that's equal to the original line's slope (m), and any perpedicular line has a slope that is the negative reciprocal (-1/m) of the original line.

Going to your problem: the slope of your original line is -1/3, so the slope of the parallel line is also -1/3.

Since you're also given a point (x1,y1) = (6,5), you can use the point-slope form of linear equations, which says that

(y - y1) = m (x - x1).

So, plug in all the values:

y - 5 = (-1/3) (x - 6)
y - 5 = (-1/3)x + 2
y = (-1/3)x + 7

Answer 3

The slope of the given line = -(1/3)
So the slope of the line parallel to given line = -(1/3)
The equation of required line, y = (-1/3)x + c
Substituting the x and y coordinates
5 = (-1/3)6 + c
c = 5 + 2= 7
So the equation of required line
y = (-1/3)x + 7
rearranging
3y = -x +21
x + 3y -21 =0

Answer 4

the coordinate is (6 ; 5) this means x=6; y=5

then, if the line is parallel to y= -1/3 x - 7

the slope should be the same = -1/3

Now using this info:

5 = -1/3 * 6 + b

5= -2 + b

b= 7

So, y= -1/3 x +7

you can verify this answer by drawing both equations together

Answer 5

y=-1/3x+7